INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Further Trig Algebraic Operations Quadratic Functions EXIT.

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Further Trig Algebraic Operations Quadratic Functions EXIT

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Algebraic Operations You have chosen to study: Please choose a question to attempt from the following: 1234 EXIT Back to Unit 3 Menu 5678

Express (m  -7) as a single fraction in its simplest form. ALGEBRAIC OPERATIONS : Question 1 Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint What would you like to do now?

Express (m  -7) as a single fraction in its simplest form. ALGEBRAIC OPERATIONS : Question 1 EXIT Use the pattern a b + c d ad + bc bd = Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only What would you like to do now?

Express (m  -7) as a single fraction in its simplest form. ALGEBRAIC OPERATIONS : Question 1 EXIT m 2 - m 2m + 14 = Go to full solution Go to Comments Go to Algebraic Ops Menu Try another like this What would you like to do now?

Comments Begin Solution Question 1 Menu Back to Home Express (m  -7) as a single fraction in its simplest form. 1. Use the pattern a b + c d ad + bc bd = = m (m + 7)- (2 x 4m) 2 (m + 7) m 2 + 7m - 8m 2m + 14 = m 2 - m 2m + 14 = Try another like this

Comments Menu Back to Home 1. Use the pattern a b + c d ad + bc bd = m (m + 7)- (2 x 4m) 2 (m + 7) m 2 + 7m - 8m 2m + 14 = m 2 - m 2m + 14 = To add or subtract fractions use the results: a b + c d ad + bc bd = a b - c d ad - bc bd = Try another

ALGEBRAIC OPERATIONS: Question 1B Express (t  3) as a single fraction in its simplest form. Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint What would you like to do now?

ALGEBRAIC OPERATIONS: Question 1B Express (t  3) as a single fraction in its simplest form. Use the pattern a b + c d ad + bc bd = Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT What would you like to do now?

ALGEBRAIC OPERATIONS: Question 1B Express (t  3) as a single fraction in its simplest form. 7t 2 + 19t 10t – 30 = Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT What would you like to do now?

Question 1B Express (t  3) as a single fraction in its simplest form. 1. Use the pattern a b + c d ad + bc bd = = 7t (t – 3)+ (10 x 4t) 10 (t – 3) 7t 2 – 21t +40t 10t – 30 = 7t 2 + 19t 10t – 30 = Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments To add or subtract fractions use the results: a b + c d ad + bc bd = a b - c d ad - bc bd = 1. Use the pattern a b + c d ad + bc bd = = 7t (t – 3)+ (10 x 4t) 10 (t – 3) 7t 2 – 21t +40t 10t – 30 = 7t 2 + 19t 10t – 30 = Menu Back to Home Next Comment

Comments 7t 2 + 19t 10t - 30 Note: Always check that you have cancelled as far as possible. This is the final result, it does not cancel further. 1. Use the pattern a b + c d ad + bc bd = = 7t (t – 3)+ (10 x 4t) 10 (t – 3) 7t 2 – 21t +40t 10t – 30 = 7t 2 + 19t 10t – 30 = Menu Back to Home

ALGEBRAIC OPERATIONS: Question 2 Express as a single fraction in its simplest form. Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint What would you like to do now?

ALGEBRAIC OPERATIONS: Question 2 Express as a single fraction in its simplest form. Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT What would you like to do now? Multiply top line then bottom line. Cancel numbers then letters in alphabetical order.

ALGEBRAIC OPERATIONS: Question 2 Express as a single fraction in its simplest form. Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT Try another like this What would you like to do now? 3b 2a =

Question 2 Express as a single fraction in its simplest form. 2a 2 5b x 15b 2 4a 3 30a 2 b 2 20a 3 b = 1. Multiply top line then bottom line. 2. Cancel numbers then letters in alphabetical order. 3 2a b Comments Begin Solution Menu Back to Home Try another like this 3b 2a = What would you like to do now?

Comments 2a 2 5b x 15b 2 4a 3 30a 2 b 2 20a 3 b = 3b 2a = 1. Multiply top line then bottom line. 2. Cancel numbers then letters in alphabetical order. 3 2a b To multiply fractions use the result: a b x c d ac bd = Menu Back to Home Next Comment

Comments 2a 2 5b x 15b 2 4a 3 30a 2 b 2 20a 3 b = 3b 2a = 1. Multiply top line then bottom line. 2. Cancel numbers then letters in alphabetical order. 3 2a b To simplify final answer write out in full and cancel: 30a 2 b 2 20a 3 b = = 3b 2a 30.a.a.b.b 20.a.a.a.b Menu Back to Home Try another

ALGEBRAIC OPERATIONS: Question 2B Express as a single fraction in its simplest form. Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint What would you like to do now?

ALGEBRAIC OPERATIONS: Question 2B Express as a single fraction in its simplest form. Multiply top line then bottom line. Cancel numbers then letters in alphabetical order. To divide by a fraction : turn it upside down and multiply. Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT What would you like to do now?

ALGEBRAIC OPERATIONS: Question 2B Express as a single fraction in its simplest form. vw 2 3 = Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT What would you like to do now?

Question 2B Express as a single fraction in its simplest form. 1.To divide by a fraction : turn it upside down and multiply. 3. Cancel numbers then letters in alphabetical order. 1 3 w2w2 v 2v 2 w  6v w3w3 2v 2 w x 6v w3w3 = 2. Multiply top line then bottom line. 2v 2 w 3 6vw = vw 2 3 = Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments To divide fractions use the result: 1.To divide by a fraction : turn it upside down and multiply. 3. Cancel numbers then letters in alphabetical order. 2v 2 w  6v w3w3 2v 2 w x 6v w3w3 = 2. Multiply top line then bottom line. 2v 2 w 3 6vw = vw 2 3 = 1 3 w2w2 v a b ÷ c d ad bc = a b x d c = Menu Back to Home Next Comment

Comments 1.To divide by a fraction : turn it upside down and multiply. 3. Cancel numbers then letters in alphabetical order. 2v 2 w  6v w3w3 2v 2 w x 6v w3w3 = 2. Multiply top line then bottom line. 2v 2 w 3 6vw = vw 2 3 = 1 3 w2w2 v To simplify final answer write out in full and cancel: 2v 2 w 3 6vw = 2.v.v.w.w.w 6.v.w vw 2 3 = Menu Back to Home Next Comment

ALGEBRAIC OPERATIONS: Question 3 Simplify Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint What would you like to do now?

Deal with numbers and then apply laws of indices: when dividing subtract the powers. Remember subtracting negative is like adding. ALGEBRAIC OPERATIONS: Question 3 Simplify Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT What would you like to do now?

ALGEBRAIC OPERATIONS: Question 3 Simplify = 3d 2 Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT What would you like to do now? Try another like this

Simplify. Question 3 1. Deal with numbers and then apply laws of indices: when dividing subtract the powers. 2. Remember subtracting negative is like adding. 24d 5 / 4  8d –3 / 4 = 3d 5 / 4 -( –3 / 4 ) = 3d 8 / 4 = 3d 2 Comments Begin Solution Menu Back to Home Try another like this

Comments 1. Deal with numbers and then apply laws of indices: when dividing subtract the powers. 2. Remember subtracting negative is like adding. 24d 5 / 4  8d –3 / 4 = 3d 5 / 4 -( –3 / 4 ) = 3d 8 / 4 = 3d 2 Learn Laws of Indices: e.g. Menu Back to Home Try another

ALGEBRAIC OPERATIONS: Question 3B Simplify Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint What would you like to do now?

Deal with top row first. Apply laws of indices: when dividing subtract the powers & when multiplying add powers. Now divide remembering subtracting negative is like adding. ALGEBRAIC OPERATIONS: Question 3B Simplify Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT What would you like to do now?

ALGEBRAIC OPERATIONS: Question 3B Simplify = a 3 Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT What would you like to do now?

Simplify. Question 3B 1. Deal with top row first. Apply laws of indices: when dividing subtract the powers & when multiplying add powers. 2. Now divide remembering subtracting negative is like adding. a -2 a 5 / 3 x a –2 / 3 = a 1-(-2) = a 5 / 3 –2 / 3 a -2 = a 3 / 3 a -2 = a 3 Comments Begin Solution Continue Solution Menu Back to Home

Comments Learn Laws of Indices: 1. Deal with top row first. Apply laws of indices: when dividing subtract the powers & when multiplying add powers. 2. Now divide remembering subtracting negative is like adding. a -2 a 5 / 3 x a –2 / 3 = a 1-(-2) = a 5 / 3 –2 / 3 a -2 = a 3 / 3 a -2 = a 3 e.g. Menu Back to Home Next Comment

ALGEBRAIC OPERATIONS: Question 4 Simplify giving your answer with positive indices. Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint What would you like to do now?

ALGEBRAIC OPERATIONS: Question 4 Simplify giving your answer with positive indices. Deal with numbers and then apply laws of indices: when dividing subtract the powers. Remember subtracting negative is like adding. Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT What would you like to do now?

ALGEBRAIC OPERATIONS: Question 4 Simplify giving your answer with positive indices. = m 2 + 1 m Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT What would you like to do now? Try another like this

Question 4 1. Multiply out brackets remembering to apply laws of indices: when multiplying add the powers. 2. Negative powers become positive on bottom line. Simplify giving your answer with positive indices. m 1 / 3 ( m 5 / 3 + m -4 / 3 ) = m 6 / 3 + m -3 / 3 = m 2 + m -1 = m 2 + 1 m Comments Begin Solution Menu Back to Home Try another like this What would you like to do now?

Comments Learn Laws of Indices: e.g. 1. Multiply out brackets remembering to apply laws of indices: when multiplying add the powers. 2. Negative powers become positive on bottom line. m 1 / 3 ( m 5 / 3 + m -4 / 3 ) = m 6 / 3 + m -3 / 3 = m 2 + m -1 = m 2 + 1 m Menu Back to Home Try another

ALGEBRAIC OPERATIONS: Question 4B Simplify giving your answer without indices. Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint What would you like to do now?

Simplify giving your answer without indices. Multiply out brackets remembering to apply laws of indices: when multiplying add the powers. Zero power is 1 and ½ power is square root. ALGEBRAIC OPERATIONS: Question 4B Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT What would you like to do now?

ALGEBRAIC OPERATIONS: Question 4B Simplify giving your answer without indices. = 1 -  w Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT What would you like to do now?

Question 4B 1. Multiply out brackets remembering to apply laws of indices: when multiplying add the powers. 2. Zero power is 1 and ½ power is square root. Simplify giving your answer without indices. w -1 / 4 ( w 1 / 4 - w 3 / 4 ) = w 0 - w 1 / 2 = 1 -  w = w -1 / 4 + 1 / 4 - w -1 / 4 + 3 / 4 Comments Begin Solution Continue Solution Menu Back to Home

Comments Learn Laws of Indices: e.g. 1. Multiply out brackets remembering to apply laws of indices: when multiplying add the powers. 2. Zero power is 1 and ½ power is square root. w -1 / 4 ( w 1 / 4 - w 3 / 4 ) = w 0 - w 1 / 2 = 1 -  w = w -1 / 4 + 1 / 4 - w -1 / 4 + 3 / 4 Menu Back to Home Next Comment

ALGEBRAIC OPERATIONS: Question 5 Evaluate7c 3 / 4 when c = 16 What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint

Deal with indices first. Power ¾ is 4th root cubed. Evaluate root before power ALGEBRAIC OPERATIONS: Question 5 Evaluate7c 3 / 4 when c = 16 What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 5 Evaluate7c 3 / 4 when c = 16 = 56 What would you like to do now? Try another like this Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 5 1.Deal with indices first. Power ¾ is 4th root cubed. = 56 Evaluate 7c 3 / 4 when c = 16 c3/4c3/4 = ( 4  c) 3 = ( 4  16) 3 = (2) 3 = 8 So 7c 3 / 4 = 7 x 8 Comments Begin Solution Menu Back to Home Try another like this What would you like to do now?

Comments Learn Laws of Indices: 1.Deal with indices first. Power ¾ is 4th root cubed. = 56 = ( 4  c) 3 = ( 4  16) 3 = (2) 3 = 8 So 7c 3 / 4 = 7 x 8 Think “Flower Power”: Power on top Root at bottom Menu Back to Home Next Comment

Comments Learn Laws of Indices: 1.Deal with indices first. Power ¾ is 4th root cubed. = 56 = ( 4  c) 3 = ( 4  16) 3 = (2) 3 = 8 So 7c 3 / 4 = 7 x 8 e.g. Always find root before power Menu Back to Home Try another

ALGEBRAIC OPERATIONS: Question 5B Evaluate10f -1 / 2 when f = 25 What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint

ALGEBRAIC OPERATIONS: Question 5B Evaluate10f -1 / 2 when f = 25 Deal with indices first. Power – ½ is square root on bottom line. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 5B = 2 Evaluate10f -1 / 2 when f = 25 What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 5B Evaluate 10f -1 / 2 when f = 25 1.Deal with indices first. Power – ½ is square root on bottom line. 10f -1 / 2 = 10 f f  25 = 10 5 = 2 Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments Learn Laws of Indices: e.g. 1.Deal with indices first. Power – ½ is square root on bottom line. 10f -1 / 2 = 10 f f  25 = 10 5 = 2 Menu Back to Home Next Comment

ALGEBRAIC OPERATIONS: Question 6 f(x) = 7x 1 / 2. Find the value of x if f(x) = 63. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint What would you like to do now?

Equate both things that are equal to f(x). To get rid of square roots square both sides. ALGEBRAIC OPERATIONS: Question 6 f(x) = 7x 1 / 2. Find the value of x if f(x) = 63. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 6 X = 81 f(x) = 7x 1 / 2. Find the value of x if f(x) = 63. What would you like to do now? Try another like this Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 6 1. Equate both things that are equal to f(x). f(x) = 7x 1 / 2. Find the value of x if f(x) = 63. 7x 1 / 2 = 63 so x 1 / 2 = 9 (  7) 2. Remember power ½ is the square root. so  x = 9 3. Now square each side. so x = 9 2 ie x = 81 Comments Begin Solution Menu Back to Home Try another like this What would you like to do now?

Comments Learn Laws of Indices: 1. Equate both things that are equal to f(x). 7x 1 / 2 = 63 so x 1 / 2 = 9 (  7) 2. Remember power ½ is the square root. so  x = 9 3. Now square each side. so x = 9 2 ie x = 81 Note: In solving the equation Square both sides Menu Back to Home Try another

ALGEBRAIC OPERATIONS: Question 6B f(x) = 2x 1 / 3. Find the value of x if f(x) = 20. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint

Equate both things that are equal to f(x). To get rid of cube roots cube both sides. ALGEBRAIC OPERATIONS: Question 6B f(x) = 2x 1 / 3. Find the value of x if f(x) = 20. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 6B X = 1000 f(x) = 2x 1 / 3. Find the value of x if f(x) = 20. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 6B 1. Equate both things that are equal to f(x). f(x) = 2x 1 / 3. Find the value of x if f(x) = 20. 2. Remember power 1/3 is the cube root. 3. Now cube each side. ie x = 1000 2x 1 / 3 = 20 so x 1 / 3 = 10 (  2) so 3  x = 10 so x = 10 3 Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments Learn Laws of Indices: Note: In solving the equation cube both sides 1. Equate both things that are equal to f(x). 2. Remember power 1/3 is the cube root. 3. Now cube each side. ie x = 1000 2x 1 / 3 = 20 so x 1 / 3 = 10 (  2) so 3  x = 10 so x = 10 3 Menu Back to Home Next Comment

ALGEBRAIC OPERATIONS: Question 7 Simplify  75 -  27 +  48 What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint

Find the largest “perfect square” factor of each of the numbers! Collect identical surds in same way as you would letters.. ALGEBRAIC OPERATIONS: Question 7 Simplify  75 -  27 +  48 What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 7 Simplify  75 -  27 +  48 = 6  3 What would you like to do now? Try another like this Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 7 1. Find the largest “perfect square” factor of each of the numbers! 2. Collect identical surds in same way as you would letters. Simplify  75 -  27 +  48 =  25 x  3 -  9 x  3 +  16 x  3 = 5  3 - 3  3 + 4  3 = 6  3 5 x – 3 x + 4 x = 6 x  75 -  27 +  48 Comments Begin Solution Menu Back to Home Try another like this

Comments Learn Laws of Indices: 1. Find the largest “perfect square” factor of each of the numbers! 2. Collect identical surds in same way as you would letters. =  25 x  3 -  9 x  3 +  16 x  3 = 5  3 - 3  3 + 4  3 = 6  3 5 x – 3 x + 4 x = 6 x e.g. Menu Back to Home Next Comment

Comments 1. Find the largest “perfect square” factor of each of the numbers! 2. Collect identical surds in same way as you would letters. =  25 x  3 -  9 x  3 +  16 x  3 = 5  3 - 3  3 + 4  3 = 6  3 5 x – 3 x + 4 x = 6 x The key to these simplification questions is that all of the individual terms can be reduced to a multiple of the same basic surd. So check that once you have taken “perfect squares” all terms have the same basic surd. If not you may not have used the highest perfect square for a term. Menu Back to Home Try another

ALGEBRAIC OPERATIONS: Question 7B Simplify  80 +  45 -  180 What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint

ALGEBRAIC OPERATIONS: Question 7B Simplify  80 +  45 -  180 Find the largest “perfect square” factor of each of the numbers! Collect identical surds in same way as you would letters.. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 7B Simplify  80 +  45 -  180 =  5 What would you like to do now? Try another like this Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 7B Simplify  80 +  45 -  180 1. Find the largest “perfect square” factor of each of the numbers! 2. Collect identical surds in same way as you would letters. =  16 x  5 +  80 +  45 -  180  9 x  5 -  36 x  5 = 4  5 + 3  5 - 6  5 =  5 4 x + 3 x – 6 x = x Comments Begin Solution Menu Back to Home Try another like this What would you like to do now?

Comments Learn Laws of Indices: 1. Find the largest “perfect square” factor of each of the numbers! 2. Collect identical surds in same way as you would letters. =  16 x  5 +  80 +  45 -  180  9 x  5 -  36 x  5 = 4  5 + 3  5 - 6  5 =  5 4 x + 3 x – 6 x = x e.g. Menu Back to Home Next Comment

Comments 1. Find the largest “perfect square” factor of each of the numbers! 2. Collect identical surds in same way as you would letters. =  16 x  5 +  80 +  45 -  180  9 x  5 -  36 x  5 = 4  5 + 3  5 - 6  5 =  5 4 x + 3 x – 6 x = x The key to these simplification questions is that all of the individual terms can be reduced to a multiple of the same basic surd. So check that once you have taken “perfect squares” all terms have the same basic surd. If not you may not have used the highest perfect square for a term. Menu Back to Home Try another

ALGEBRAIC OPERATIONS: Question 7C Simplify  75 -  27 +  48  300 -  12 What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint

Deal with top and bottom lines separately. For each, find the largest “perfect square” factor of each of the numbers! ALGEBRAIC OPERATIONS: Question 7C Simplify  75 -  27 +  48  300 -  12 Now bring both parts together again. Cancel out where the same basic surd is in evidence on both top and bottom. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 7C Simplify  75 -  27 +  48  300 -  12 3/43/4 = What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 7C 1. Deal with top and bottom lines separately. For each, find the largest “perfect square” factor of each of the numbers! 5 x - 3 x + 4 x = 6 x Simplify  75 -  27 +  48  300 -  12 =  25 x  3 -  75 -  27 +  48 = 5  3 - 3  3 + 4  3 = 6  3 Top Line:  9 x  3 +  16 x  3 Comments Begin Solution Continue Solution Menu Back to Home

Question 7C 1. Deal with top and bottom lines seperately. For each, find the largest “perfect square” factor of each of the numbers! 10 x - 2 x = 8 x Simplify  75 -  27 +  48  300 -  12 Bottom Line:  300 -  12 =  100 x  3 - = 10  3 - 2  3 = 8  3  4 x  3 Comments Begin Solution Continue Solution Menu Back to Home

Question 7C 2. Now bring both parts together again. Cancel out where the same basic surd is in evidence on both top and bottom. Simplify  75 -  27 +  48  300 -  12  75 -  27 +  48  300 -  12 6363 8383 3/43/4 = = 3 4 Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments Learn Laws of Indices: e.g. 1. Deal with top and bottom lines seperately. For each, find the largest “perfect square” factor of each of the numbers! 5 x - 3 x + 4 x = 6 x =  25 x  3 -  75 -  27 +  48 = 5  3 - 3  3 + 4  3 = 6  3 Top Line:  9 x  3 +  16 x  3 Menu Back to Home Next Comment

Comments In questions involving fractions always treat each line separately. Reduce each line to its simplest form before dividing. 2. Now bring both parts together again. Cancel out where the same basic surd is in evidence on both top and bottom.  75 -  27 +  48  300 -  12 6363 8383 3/43/4 = = The key to these simplification questions is that all of the individual terms can be reduced to a multiple of the same basic surd. In fractions these will then often cancel. Remember to cancel numbers too!! Menu Back to Home Next Comment

ALGEBRAIC OPERATIONS: Question 8 12 66 Expresswith a rational denominator. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint

To change the look of a fraction but not the value multiply top & bottom by the same amount ie  6. ALGEBRAIC OPERATIONS: Question 8 12 66 Express with a rational denominator. What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 8 = 2  6 12 66 Expresswith a rational denominator. What would you like to do now? Try another like this Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 8 1. To change the look of a fraction but not the value multiply top & bottom by the same amount ie  6. 12 66 Express with a rational denominator. 12 66 = 66 x 66 66 12  6 6 = = 2  6 2 1 Comments Begin Solution Menu Back to Home Try another like this

Comments Learn Laws of Surds: Rationalising the denominator: Required to remove from the denominator. Multiply top and bottom by 1. To change the look of a fraction but not the value multiply top & bottom by the same amount ie  6. 12 66 = 66 x 66 66 12  6 6 = = 2  6 Menu Back to Home Next Comment

Comments Learn Laws of Surds: e.g. Rationalising the denominator: 1. To change the look of a fraction but not the value multiply top & bottom by the same amount ie  6. 12 66 = 66 x 66 66 12  6 6 = = 2  6 Menu Back to Home Try another

ALGEBRAIC OPERATIONS: Question 8B  84 33 Simplify What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint

ALGEBRAIC OPERATIONS: Question 8B Find the largest “perfect square” factor of each of the numbers! Collect identical surds in same way as you would letters..  84 33 Simplify What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 8B = 2  7  84 33 Simplify What would you like to do now? Try another like this Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 8B 1. Re-write expression using laws of surds  84 33 Simplify = 2  7  84 33 = 84 3  =  28 =  4 x  7 Comments Begin Solution Menu Back to Home Try another like this What would you like to do now?

Comments Learn Laws of Surds: e.g. 1. Re-write expression using laws of surds = 2  7  84 33 = 84 3  =  28 =  4 x  7 Menu Back to Home Next Comment

Comments Learn Laws of Surds: e.g. 1. Re-write expression using laws of surds = 2  7  84 33 = 84 3  =  28 =  4 x  7 Menu Back to Home Try another

ALGEBRAIC OPERATIONS: Question 8C Find the value of tanx° giving your answer as a fraction with a rational denominator. x°x° 3m 6  2m What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT Get hint

To change the look of a fraction but not the value multiply top & bottom by the same amount ie  2. ALGEBRAIC OPERATIONS: Question 8C When dealing with fractions check that you have simplified as far as possible. Find the value of tanx° giving your answer as a fraction with a rational denominator. x°x° 3m 6  2m What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu Reveal answer only EXIT

ALGEBRAIC OPERATIONS: Question 8C Find the value of tanx° giving your answer as a fraction with a rational denominator. x°x° 3m 6  2m 22 4 = What would you like to do now? Go to full solution Go to Comments Go to Algebraic Ops Menu EXIT

Question 8C Find the value of tanx° giving your answer as a fraction with a rational denominator. x°x° 3m 6  2m 1. To change the look of a fraction but not the value multiply top & bottom by the same amount ie  2. tanx° =3 6262 = 3 6262 x 22 22 3232 6x2 = 3232 12 = 22 4 = Simplify!!! 1 4 Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments Learn Laws of Surds: 1. To change the look of a fraction but not the value multiply top & bottom by the same amount ie  2. tanx° =3 6262 = 3 6262 x 22 22 3232 6x2 = 3232 12 = 22 4 = Rationalising the denominator: Required to remove from the denominator. Multiply top and bottom by Menu Back to Home Next Comment

Comments Learn Laws of Surds: e.g. 1. To change the look of a fraction but not the value multiply top & bottom by the same amount ie  2. tanx° =3 6262 = 3 6262 x 22 22 3232 6x2 = 3232 12 = 22 4 = In questions involving fractions make sure you have simplified as far as possible. Simplify!!! Menu Back to Home Next Comment

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Further Trig You have chosen to study: Please choose a question to attempt from the following: 12 EXIT Back to Unit 3 Menu

Further Trig : Question 1 The graph below shows a curve with equation in the form y = asinbx°. Write down the values of a and b. 120 0 240 0 360 0 y = asinbx° EXIT Go to full solution Go to Comments Further Trig Menu Reveal answer only Get hint What would you like to do now?

Further Trig : Question 1 The graph below shows a curve with equation in the form y = asinbx°. Write down the values of a and b. 120 0 240 0 360 0 y = asinbx° a = amplitude = ½ vertical extent. b = no. of waves in 360 0 EXIT Go to full solution Go to Comments Further Trig Menu Reveal answer only What would you like to do now?

Further Trig : Question 1 The graph below shows a curve with equation in the form y = asinbx°. Write down the values of a and b. 120 0 240 0 360 0 y = asinbx° a = 2 b = 3 EXIT Go to full solution Go to Comments Further Trig Menu What would you like to do now? Try another like this

Question 1 120 0 240 0 360 0 y = asinbx° Write down the values of a and b. 1. a = amplitude = ½ vertical extent. max /min = ±2 soa = 2 2. b = no. of waves in 360 0 3 complete waves from 0 to 360 so b = 3 Comments Begin Solution Menu Back to Home Try another like this What would you like to do now?

Comments 1. a = amplitude = ½ vertical extent. max /min = ±2 soa = 2 2. b = no. of waves in 360 0 3 complete waves from 0 to 360 so b = 3 Learn basic trig graphs: i.e.y = sinx˚ Max. = 1 Min. = -1 One cycle in 360˚ Menu Back to Home Next Comment

Comments 1. a = amplitude = ½ vertical extent. max /min = ±2 soa = 2 2. b = no. of waves in 360 0 3 complete waves from 0 to 360 so b = 3 Learn basic trig graphs: i.e.y = sinx˚ y = asinx˚ Stretch factor y = sinbx˚ Number of cycles in 360˚ One cycle Menu Back to Home Try another

Further Trig : Question 1B The graph below shows a curve with equation in the form y = acosbx°. Write down the values of a and b. y = acosbx° 90 0 180 0 360 0 270 0 EXIT Go to full solution Go to Comments Further Trig Menu Reveal answer only Get hint What would you like to do now?

Further Trig : Question 1B The graph below shows a curve with equation in the form y = acosbx°. Write down the values of a and b. y = acosbx° 90 0 180 0 360 0 270 0 a = amplitude = ½ vertical extent. b = no. of waves in 360 0 EXIT Go to full solution Go to Comments Further Trig Menu Reveal answer only What would you like to do now?

Further Trig : Question 1B The graph below shows a curve with equation in the form y = acosbx°. Write down the values of a and b. y = acosbx° 90 0 180 0 360 0 270 0 a = 5 b = 2 EXIT Go to full solution Go to Comments Further Trig Menu What would you like to do now?

y = acosbx° 90 0 180 0 360 0 270 0 Question 1B Write down the values of a and b. 1. a = amplitude = ½ vertical extent. max /min = ±5 soa = 5 2. b = no. of waves in 360 0 2 complete waves from 0 to 360 so b = 2 Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments 1. a = amplitude = ½ vertical extent. max /min = ±5 soa = 5 2. b = no. of waves in 360 0 2 complete waves from 0 to 360 so b = 2 Learn basic trig graphs: i.e.y = cosx˚ Max. = 1 Min. = -1 One cycle in 360˚ Menu Back to Home Next Comment

Comments 1. a = amplitude = ½ vertical extent. max /min = ±2 soa = 2 2. b = no. of waves in 360 0 3 complete waves from 0 to 360 so b = 3 Learn basic trig graphs: i.e.y = cosx˚ y = acosx˚ Stretch factor y = cosbx˚ Number of cycles in 360˚ One cycle Menu Back to Home

Further Trig : Question 2 Solve 5sin x ° - 1 = 1where 0<x<360 EXIT Go to full solution Go to Comments Further Trig Menu Reveal answer only Get hint What would you like to do now?

Further Trig : Question 2 Solve 5sin x ° - 1 = 1where 0<x<360 Always re-arrange given equation to: sin x = cosx = tan x= Use CAST diagram to decide which quadrants angles lie. Remember that answers to trig equations come in pairs. EXIT Go to full solution Go to Comments Further Trig Menu Reveal answer only What would you like to do now?

Further Trig : Question 2 Solve 5sin x ° - 1 = 1where 0<x<360 x = 23.6° x = 156.4° EXIT Go to full solution Go to Comments Further Trig Menu What would you like to do now? Try another like this

Question 2 1. Always re-arrange to sinx 0 = …... 2. Use the CAST diagram to decide which quadrant angles lie in. Solve 5sin x ° - 1 = 1 where 0<x<360 sinx  = 0.4 5sinx° - 1 = 1 5sinx° = 2 180 -   180 +  360 -  sin all tan cos Q1 or Q2 Where is sin positive? Comments Begin Solution Continue Solution Menu Back to Home

Question 2 1. Always re-arrange to sinx 0 = …... 3. Calculate angles remembering that answers to trig equations come in pairs. Solve 5sin x ° - 1 = 1 where 0<x<360 sinx  = 0.4 5sinx° - 1 = 1 5sinx° = 2 Q1 or Q2 sin -1 0.4 = 23.6  Q1: angle = 23.6° Q2: angle = 180° - 23.6° = 156.4° Comments Begin Solution Menu Back to Home Try another like this What would you like to do now?

Comments We are finding the two angles at which the sine curve reaches a height of 0.4 0.4 1. Always re-arrange to sinx 0 = …... 3. Calculate angles remembering that answers to trig equations come in pairs. sinx  = 0.4 5sinx° - 1 = 1 5sinx° = 2 Q1 or Q2 Q2: angle = 180° - 23.6° sin -1 0.4 = 23.6  Q1: angle = 23.6° = 156.4° Menu Back to Home Try another

Further Trig : Question 2B Solve 3tan x ° + 7 = 2where 0<x<360 EXIT Go to full solution Go to Comments Further Trig Menu Reveal answer only Get hint What would you like to do now?

Further Trig : Question 2B Solve 3tan x ° + 7 = 2where 0<x<360 Always re-arrange given equation to: sin x = cosx = tan x= Use CAST diagram to decide which quadrants angles lie. Remember that answers to trig equations come in pairs. EXIT Go to full solution Go to Comments Further Trig Menu Reveal answer only What would you like to do now?

Solve 3tan x ° + 7 = 2where 0<x<360 Further Trig : Question 2B x = 121.0° x = 301.0° EXIT Go to full solution Go to Comments Further Trig Menu What would you like to do now?

Question 2B Solve 3tan x ° + 7 = 2 where 0<x<360 1. Always re-arrange to tanx 0 = …... tanx  = -5 / 3 3tanx° + 7 = 2 3tanx° = -5 2. Use the CAST diagram to decide which quadrant angles lie in. 180 -   180 +  360 -  sin all tan cos Q2 or Q4 Where is tan negative? Comments Begin Solution Continue Solution Menu Back to Home

Question 2B Solve 3tan x ° + 7 = 2 where 0<x<360 1. Always re-arrange to tanx 0 = …... tanx  = -5 / 3 3tanx° + 7 = 2 3tanx° = -5 Q2 or Q4 tan -1 (5  3) = 59.0  Q2: angle = 180° - 59.0° Q4: angle = 360° - 59.0° 3. Calculate angles remembering that answers to trig equations come in pairs. = 301.0° = 121.0° Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments 1. Always re-arrange to tanx 0 = …... tanx  = -5 / 3 3tanx° + 7 = 2 3tanx° = -5 tan -1 (5  3) = 59.0  Q2: angle = 180° - 59.0° Q4: angle = 360° - 59.0° 3. Calculate angles remembering that answers to trig equations come in pairs. = 301.0° = 121.0° We are finding the two angles at which the tan graph reaches a height of Menu Back to Home Next Comment

Comments 1. Always re-arrange to tanx 0 = …... tanx  = -5 / 3 3tanx° + 7 = 2 3tanx° = -5 tan -1 (5  3) = 59.0  Q2: angle = 180° - 59.0° Q4: angle = 360° - 59.0° 3. Calculate angles remembering that answers to trig equations come in pairs. = 301.0° = 121.0° Note: Never put a negative value into the inverse trig function DO: then CAST diagram Menu Back to Home Next Comment

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Quadratic Functions You have chosen to study: Please choose a question to attempt from the following: 12 EXIT Back to Unit 3 Menu 34

Quadratic Functions: Question 1 Solve the equation x 2 – 6x + 7 = 0 giving your answers to 2 decimal places. EXIT Go to full solution Go to Comments Go to Quadratics Menu Reveal answer only Get hint What would you like to do now?

Quadratic Functions: Question 1 Solve the equation x 2 – 6x + 7 = 0 giving your answers to 2 decimal places. Write down values of a, b & c. Evaluate b 2 – 4ac. Now use quadratic formula, rememberin g that you should get two answers. Remember to round. EXIT Go to full solution Go to Comments Go to Quadratics Menu Reveal answer only What would you like to do now?

Quadratic Functions: Question 1 Solve the equation x 2 – 6x + 7 = 0 giving your answers to 2 decimal places. x = 4.41 or 1.59 EXIT Go to full solution Go to Comments Go to Quadratics Menu Try another like this What would you like to do now?

Question 1 Solve the equation x 2 – 6x + 7 = 0 giving your answers to 2 decimal places. 1. Write down values of a, b & c. x 2 – 6x + 7 = 0 a = 1 b = (-6) c = 7 2. Evaluate b 2 – 4ac. b 2 – 4ac = (-6) 2 – (4x1x7) = 36 – 28 = 8 Comments Begin Solution Continue Solution Menu Back to Home

Question 1 Solve the equation x 2 – 6x + 7 = 0 giving your answers to 2 decimal places. 1. Write down values of a, b & c. x 2 – 6x + 7 = 0 a = 1 b = -6 c = 7 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = 6 ±  8 2 4. Rewrite with brackets and now use calculator. = (6 +  8)  2 or (6 -  8)  2 Comments Begin Solution Continue Solution Menu Back to Home

Question 1 Solve the equation x 2 – 6x + 7 = 0 giving your answers to 2 decimal places. 1. Write down values of a, b & c. x 2 – 6x + 7 = 0 a = 1 b = -6 c = 7 4. Rewrite with brackets and now use calculator. = (6 +  8)  2 or (6 -  8)  2 = 4.414… or 1.585.. 5. Remember to round. x = 4.41 or 1.59 Comments Begin Solution Try another like this Menu Back to Home What would you like to do now?

Comments 1. Write down values of a, b & c. x 2 – 6x + 7 = 0 a = 1 b = -6 c = 7 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = 6 ±  8 2 4. Rewrite with brackets and now use calculator. = (6 +  8)  2 or (6 -  8)  2 For any question involving a quadratic in which you are asked to give your answer to a given number of decimal places OR a given number of significant figures: THINK QUADRATIC FORMULA!! Menu Back to Home Next Comment

Comments 1. Write down values of a, b & c. x 2 – 6x + 7 = 0 a = 1 b = -6 c = 7 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = 6 ±  8 2 4. Rewrite with brackets and now use calculator. = (6 +  8)  2 or (6 -  8)  2 ax 2 + bx + c = 0 Refer to the Formula Sheet: x = -b ±  (b 2 – 4ac ) 2a Quadratic Formula: Menu Back to Home Next Comment

Comments 1. Write down values of a, b & c. x 2 – 6x + 7 = 0 a = 1 b = -6 c = 7 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = 6 ±  8 2 4. Rewrite with brackets and now use calculator. = (6 +  8)  2 or (6 -  8)  2 Take care when allocating a value to a, b and c. ax 2 + bx + c = 0 1x 2 - 6x + 7 = 0 a =1 b = (-6) c = 7 Menu Back to Home Next Comment

Note: Finding first eases working. Bracket all negative numbers. Watch out for double negative! Comments 1. Write down values of a, b & c. x 2 – 6x + 7 = 0 a = 1 b = -6 c = 7 b 2 – 4ac 2. Evaluate b 2 – 4ac. b 2 – 4ac = (-6) 2 – (4x1x7) = 36 – 28 = 8 Menu Back to Home Try another

Quadratic Functions: Question 1B Go to full solution Go to Comments Go to Quadratics Menu Reveal answer only EXIT Get hint Solve the equation 3x 2 + 5x - 1 = 0 giving your answers to 2 significant figures. What would you like to do now?

Quadratic Functions: Question 1B Go to full solution Go to Comments Go to Quadratics Menu Reveal answer only EXIT Write down values of a, b & c. Evaluate b 2 – 4ac. Now use quadratic formula, rememberin g that you should get two answers. Remember to round. Solve the equation 3x 2 + 5x - 1 = 0 giving your answers to 2 significant figures. What would you like to do now?

Quadratic Functions: Question 1B Go to full solution Go to Comments Go to Quadratics Menu EXIT x = 0.18 or -1.8 Solve the equation 3x 2 + 5x - 1 = 0 giving your answers to 2 significant figures. What would you like to do now?

Comments Begin Solution Continue Solution Question 1B Quadratics Menu Back to Home 1. Write down values of a, b & c. 3x 2 + 5x – 1 = 0 a = 3 b = 5 c = (-1) 2. Evaluate b 2 – 4ac. b 2 – 4ac = (5) 2 – (4 x 3 x -1) = 25 – (- 12) = 37 Solve the equation 3x 2 + 5x - 1 = 0 giving your answers to 2 significant figures.

Comments Begin Solution Continue Solution Question 1B Quadratics Menu Back to Home 1. Write down values of a, b & c. 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = -5 ±  37 2 4. Rewrite with brackets and now use calculator. = (-5 +  37)  2 or (-5 -  37)  2 3x 2 + 5x – 1 = 0 a = 3 b = 5 c = (-1) Solve the equation 3x 2 + 5x - 1 = 0 giving your answers to 2 significant figures.

Comments Begin Solution Continue Solution Question 1B Quadratics Menu Back to Home 1. Write down values of a, b & c. 4. Rewrite with brackets and now use calculator. = (-5 +  37)  2 or (-5 -  37)  2 5. Remember to round. x = 0.18 or -1.8 3x 2 + 5x – 1 = 0 a = 3 b = 5 c = (-1) Solve the equation 3x 2 + 5x - 1 = 0 giving your answers to 2 significant figures. = 0.180… or -1.847.. What would you like to do now?

Comments Quadratics Menu Back to Home Next Comment For any question involving a quadratic in which you are asked to give your answer to a given number of decimal places OR a given number of significant figures: THINK QUADRATIC FORMULA!! 1. Write down values of a, b & c. 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = -5 ±  37 2 4. Rewrite with brackets and now use calculator. = (-5 +  37)  2 or (-5 -  37)  2 3x 2 + 5x – 1 = 0 a = 3 b = 5 c = (-1)

Comments Quadratics Menu Back to Home Next Comment ax 2 + bx + c = 0 Refer to the Formula Sheet: x = -b ±  (b 2 – 4ac ) 2a Quadratic Formula: 1. Write down values of a, b & c. 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = -5 ±  37 2 4. Rewrite with brackets and now use calculator. = (-5 +  37)  2 or (-5 -  37)  2 3x 2 + 5x – 1 = 0 a = 3 b = 5 c = (-1)

Comments Quadratics Menu Back to Home Next Comment Take care when allocating a value to a, b and c. ax 2 + bx + c = 0 3x 2 + 5x - 1 = 0 a =3 b = 5 c = -1 1. Write down values of a, b & c. 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = -5 ±  37 2 4. Rewrite with brackets and now use calculator. = (-5 +  37)  2 or (-5 -  37)  2 3x 2 + 5x – 1 = 0 a = 3 b = 5 c = (-1)

Note: Finding first eases working. Bracket all negative numbers. Watch out for double negative! Comments Quadratics Menu Back to Home Next Comment b 2 – 4ac 1. Write down values of a, b & c. 3x 2 + 5x – 1 = 0 a = 3 b = 5 c = (-1) 2. Evaluate b 2 – 4ac. b 2 – 4ac = (5) 2 – (4 x 3 x -1) = 25 – (- 12) = 37

Quadratic Functions: Question 2 Go to full solution Go to Comments Go to Quadratics Menu Reveal answer only EXIT Get hint Solve the equation 3x 2 = x + 1 giving your answers to 2 decimal places. What would you like to do now?

Quadratic Functions: Question 2 Go to full solution Go to Comments Go to Quadratics Menu Reveal answer only EXIT Write down values of a, b & c. Evaluate b 2 – 4ac. Now use quadratic formula, rememberin g that you should get two answers. Remember to round. Solve the equation 3x 2 = x + 1 giving your answers to 2 decimal places. Re-write quadratic: make it equal to zero before solving. What would you like to do now?

Quadratic Functions: Question 2 Go to full solution Go to Comments Go to Quadratics Menu EXIT Try another like this x = 0.77 or -0.43 Solve the equation 3x 2 = x + 1 giving your answers to 2 decimal places. What would you like to do now?

Comments Begin Solution Question 2 Quadratics Menu Back to Home 2. Write down values of a, b & c. 3x 2 – x – 1 = 0 a = 3 b = (-1) c = (-1) 3. Evaluate b 2 – 4ac. b 2 – 4ac = (-1) 2 – (4 x 3 x -1) = 1 – (-12) = 13 Continue Solution Solve the equation 3x 2 = x + 1 giving your answers to 2 decimal places. 1. Rearrange in quadratic form. 3x 2 = x + 1

Comments Begin Solution Continue Solution Question 2 Quadratics Menu Back to Home 2. Write down values of a, b & c. 3x 2 –x – 1 = 0 a = 3 b = (-1) c = (-1) 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = 1 ±  13 2 4. Rewrite with brackets and now use calculator. = (1 +  13)  2 or (1 -  13)  2 Solve the equation 3x 2 = x + 1 giving your answers to 2 decimal places.

Comments Begin Solution Question 2 Quadratics Menu Back to Home 1. Write down values of a, b & c. 3x 2 – x – 1 = 0 a = 3 b = (-1) c = (-1) 4. Rewrite with brackets and now use calculator. = (1 +  13)  2 or (1 -  13)  2 5. Remember to round. x = 0.77 or -0.43 Try another like this Solve the equation 3x 2 = x + 1 giving your answers to 2 decimal places. = 0.767… or -0.434.. What would you like to do now?

Comments Quadratics Menu Back to Home Next Comment For any question involving a quadratic in which you are asked to give your answer to a given number of decimal places OR a given number of significant figures: THINK QUADRATIC FORMULA!! 2. Write down values of a, b & c. 3x 2 – x – 1 = 0 a = 3 b = (-1) c = (-1) 3. Evaluate b 2 – 4ac. b 2 – 4ac = (-1) 2 – (4 x 3 x -1) = 1 – (-12) = 13 1. Rearrange in quadratic form. 3x 2 = x + 1

Comments Quadratics Menu Back to Home Next Comment You must put the quadratic into standard quadratic form (make equation equal zero) before attempting to solve. 2. Write down values of a, b & c. 3x 2 – x – 1 = 0 a = 3 b = (-1) c = (-1) 3. Evaluate b 2 – 4ac. b 2 – 4ac = (-1) 2 – (4 x 3 x -1) = 1 – (-12) = 13 1. Rearrange in quadratic form. 3x 2 = x + 1

Comments Quadratics Menu Back to Home Next Comment ax 2 + bx + c = 0 Refer to the Formula Sheet: x = -b ±  (b 2 – 4ac ) 2a Quadratic Formula: 2. Write down values of a, b & c. 3x 2 –x – 1 = 0 a = 3 b = (-1) c = (-1) 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = 1 ±  13 2 4. Rewrite with brackets and now use calculator. = (1 +  13)  2 or (1 -  13)  2

Comments Quadratics Menu Back to Home Next Comment Take care when allocating a value to a, b and c. ax 2 + bx + c = 0 3x 2 - 1x - 1 = 0 a =3 b = (-1) c = (-1) 2. Write down values of a, b & c. 3x 2 –x – 1 = 0 a = 3 b = (-1) c = (-1) 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = 1 ±  13 2 4. Rewrite with brackets and now use calculator. = (1 +  13)  2 or (1 -  13)  2

Note: Finding first eases working. Bracket all negative numbers. Watch out for double negative! Comments Quadratics Menu Back to Home b 2 – 4ac 2. Write down values of a, b & c. 3x 2 – x – 1 = 0 a = 3 b = (-1) c = (-1) 3. Evaluate b 2 – 4ac. b 2 – 4ac = (-1) 2 – (4 x 3 x -1) = 1 – (-12) = 13 1. Rearrange in quadratic form. 3x 2 = x + 1 Try another like this

Quadratic Functions: Question 2B Go to full solution Go to Comments Go to Quadratics Menu Reveal answer only EXIT Get hint Solve the equation 2 x ( x + 2) = 2 - x giving your answers to 2 significant figures. What would you like to do now?

Quadratic Functions: Question 2B Go to full solution Go to Comments Go to Quadratics Menu Reveal answer only EXIT Write down values of a, b & c. Evaluate b 2 – 4ac. Now use quadratic formula, rememberin g that you should get two answers. Remember to round. Solve the equation 2 x ( x + 2) = 2 - x giving your answers to 2 significant figures. Re-write quadratic: get rid of brackets & make it equal to zero before solving. What would you like to do now?

Quadratic Functions: Question 2B Go to full solution Go to Comments Go to Quadratics Menu EXIT x = 0.35 or -2.9 Solve the equation 2 x ( x + 2) = 2 - x giving your answers to 2 significant figures. What would you like to do now?

Comments Begin Solution Continue Solution Question 2B Quadratics Menu Back to Home Solve the equation 2 x ( x + 2) = 2 - x giving your answers to 2 significant figures. 2. Write down values of a, b & c. 2x 2 + 5x – 2 = 0 a = 2 b = 5 c = (-2) 3. Evaluate b 2 – 4ac. b 2 – 4ac = (5) 2 – (4 x 2 x -2) = 25 – (-16) = 41 1. Rearrange in quadratic form. 2x(x + 2) = 2 – x 2x 2 + 4x – 2 + x= 0

Comments Begin Solution Continue Solution Question 2B Quadratics Menu Back to Home 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = -5 ±  41 2 4. Rewrite with brackets and now use calculator. = (-5 +  41)  2 or (-5 -  41)  2 Solve the equation 2 x ( x + 2) = 2 - x giving your answers to 2 significant figures. 2x 2 + 5x – 2 = 0 a = 2 b = 5 c = (-2) 1. Rearrange in quadratic form.

Comments Begin Solution Continue Solution Question 2B Quadratics Menu Back to Home 4. Rewrite with brackets and now use calculator. = (-5 +  41)  2 or (-5 -  41)  2 5. Remember to round. x = 0.35 or -2.9 2x 2 + 5x – 2 = 0 a = 2 b = 5 c = (-2) 1. Rearrange in quadratic form. Solve the equation 2 x ( x + 2) = 2 - x giving your answers to 2 significant figures. = 0.350… or -2.850.. What would you like to do now?

Comments Quadratics Menu Back to Home Next Comment For any question involving a quadratic in which you are asked to give your answer to a given number of decimal places OR a given number of significant figures: THINK QUADRATIC FORMULA!! 2. Write down values of a, b & c. 2x 2 + 5x – 2 = 0 a = 2 b = 5 c = (-2) 3. Evaluate b 2 – 4ac. b 2 – 4ac = (5) 2 – (4 x 2 x -2) = 25 – (-16) = 41 1. Rearrange in quadratic form. 2x(x + 2) = 2 – x 2x 2 + 4x – 2 + x= 0

Comments Quadratics Menu Back to Home Next Comment 2. Write down values of a, b & c. 2x 2 + 5x – 2 = 0 a = 2 b = 5 c = (-2) 3. Evaluate b 2 – 4ac. b 2 – 4ac = (5) 2 – (4 x 2 x -2) = 25 – (-16) = 41 1. Rearrange in quadratic form. 2x(x + 2) = 2 – x 2x 2 + 4x – 2 + x= 0 You must put the quadratic into standard quadratic form (make equation equal zero) before attempting to solve.

Comments Quadratics Menu Back to Home Next Comment ax 2 + bx + c = 0 Refer to the Formula Sheet: x = -b ±  (b 2 – 4ac ) 2a Quadratic Formula: 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = -5 ±  41 2 4. Rewrite with brackets and now use calculator. = (-5 +  41)  2 or (-5 -  41)  2 2x 2 + 5x – 2 = 0 a = 2 b = 5 c = (-2) 1. Rearrange in quadratic form.

Comments Quadratics Menu Back to Home Next Comment Take care when allocating a value to a, b and c. ax 2 + bx + c = 0 2x 2 + 5x - 2 = 0 a =2 b = 5 c = -2 3. Now use quadratic formula. x = -b ±  (b 2 – 4ac ) 2a x = -5 ±  41 2 4. Rewrite with brackets and now use calculator. = (-5 +  41)  2 or (-5 -  41)  2 2x 2 + 5x – 2 = 0 a = 2 b = 5 c = (-2) 1. Rearrange in quadratic form.

Note: Finding first eases working. Bracket all negative numbers. Watch out for double negative! Comments Quadratics Menu Back to Home Next Comment b 2 – 4ac 2. Write down values of a, b & c. 2x 2 + 5x – 2 = 0 a = 2 b = 5 c = (-2) 3. Evaluate b 2 – 4ac. b 2 – 4ac = (5) 2 – (4 x 2 x -2) = 25 – (-16) = 41 1. Rearrange in quadratic form. 2x(x + 2) = 2 – x 2x 2 + 4x – 2 + x= 0

Quadratic Functions: Question 3 Go to full solution Go to Comments Go to Quadratics MenuReveal answer EXIT Get hint The diagram below shows an L-shaped plot of land with dimensions as given. (x+4) m x m (x+3) m (a)Show that the total area is given by the expression x 2 + 7x m 2. (b)Hence find the value of x when this area is 60 m 2. What would you like to do now?

The diagram below shows an L-shaped plot of land with dimensions as given. (x+4) m x m (x+3) m (a)Show that the total area is given by the expression x 2 + 7x m 2. (b)Hence find the value of x when this area is 60 m 2. Quadratic Functions: Question 3 EXIT Find the area of each rectangle.. Add these to find total area. In (b) make expression from (a) = 60. Make it equal to zero, factorise & solve. Go to full solution Go to Comments Go to Quadratics Menu Reveal answer What would you like to do now?

The diagram below shows an L-shaped plot of land with dimensions as given. (x+4) m x m (x+3) m (a)Show that the total area is given by the expression x 2 + 7x m 2. (b)Hence find the value of x when this area is 60 m 2. Quadratic Functions: Question 3 EXIT Try another like this x = 5m Go to full solution Go to Comments Go to Quadratics Menu What would you like to do now?

Comments Begin Solution Question 3 Quadratics Menu Back to Home Continue Solution (x+4) m x m (x+3) m 1. Find the area of each rectangle separately and add together to get total area. Area rectangle = x(x+4) = x 2 + 4x m 2 (a) 1 2 1 Length rectangle = (x + 3) 2 = 3m – x = 3x m 2 Area rectangle 2 Hence total area = (x 2 + 4x) + 3x = x 2 + 7x m 2 (a)Show that the total area is given by the expression x 2 + 7x m 2. 3m

Comments Begin Solution Question 3 Quadratics Menu Back to Home (x+4) m x m (x+3) m 2. Use the expression for area given in part (a) and make it equal to 60. If total area = 60 (b) 1 2 And total area= x 2 + 7x (b) Hence find the value of x when this area is 60 m 2. then x 2 + 7x = 60 x 2 + 7x - 60 = 0 (x + 12)(x – 5) = 0 So (x + 12) = 0 or (x – 5) = 0 ie x = -12 or x = 5 Length must be 5m as negative value not valid. 3. Make the equation equal to zero, factorise and solve. Try another like this What would you like to do now?

Comments Quadratics Menu Back to Home Next Comment 2. Use the expression for area given in part (a) and make it equal to 60. If total area = 60 (b) And total area= x 2 + 7x then x 2 + 7x = 60 x 2 + 7x - 60 = 0 (x + 12)(x – 5) = 0 So (x + 12) = 0 or (x – 5) = 0 ie x = -12 or x = 5 Length must be 5m as negative value not valid. 3. Make the equation equal to zero, factorise and solve. The quadratic equation can also be solved by applying the quadratic formula: ax 2 + bx + c = 0 Refer to the Formula Sheet: x = -b ±  (b 2 – 4ac ) 2a Quadratic Formula: Finding first eases working. Bracket all negative numbers. b 2 – 4ac

Comments Quadratics Menu Back to Home Next Comment 2. Use the expression for area given in part (a) and make it equal to 60. If total area = 60 (b) And total area= x 2 + 7x then x 2 + 7x = 60 x 2 + 7x - 60 = 0 (x + 12)(x – 5) = 0 So (x + 12) = 0 or (x – 5) = 0 ie x = -12 or x = 5 Length must be 5m as negative value not valid. 3. Make the equation equal to zero, factorise and solve. Take care when allocating a value to a, b and c. ax 2 + bx + c = 0 1x 2 + 7x - 60 = 0 a =1 b = 7 c = (-60)

Comments Quadratics Menu Back to Home 2. Use the expression for area given in part (a) and make it equal to 60. If total area = 60 (b) And total area= x 2 + 7x then x 2 + 7x = 60 x 2 + 7x - 60 = 0 (x + 12)(x – 5) = 0 So (x + 12) = 0 or (x – 5) = 0 ie x = -12 or x = 5 Length must be 5m as negative value not valid. 3. Make the equation equal to zero, factorise and solve. Make sure that you understand when a negative answer may not be correct in context given, e.g. negative lengths or areas. Try another like this

Quadratic Functions: Question 3B Go to full solution Go to Comments Go to Quadratics Menu Reveal answer EXIT Get hint The diagram below shows plans of the foundations for a house and garage. 12m 7mhouse x m ? mgarage Planning regulations state that the length of the garage should be 4m longer than its width and its floor area should be 25% of the floor area of the house. (b)Hence find the value of x to meet this requirement. (a)Find an expression for the area of the garage in terms of x. What would you like to do now?

(b)Hence find the value of x to meet this requirement. (a)Find an expression for the area of the garage in terms of x. Quadratic Functions: Question 3B EXIT Area = length x width. Use information to find length of garage. Find the numerical value of the required area for the garage. Use expression from (a) to form a quadratic equation. Make it equal to zero, factorise & solve. Go to full solution Go to Comments Go to Quadratics Menu Reveal answer The diagram below shows plans of the foundations for a house and garage. 12m 7mhouse x m ? mgarage Planning regulations state that the length of the garage should be 4m longer than its width and its floor area should be 25% of the floor area of the house. What would you like to do now?

The diagram below shows plans of the foundations for a house and garage. 12m 7mhouse x m ? mgarage Planning regulations state that the length of the garage should be 4m longer than its width and its floor area should be 25% of the floor area of the house. (b)Hence find the value of x to meet this requirement. (a)Find an expression for the area of the garage in terms of x. Quadratic Functions: Question 3B EXIT Go to full solution Go to CommentsGo to Quadratics Menu = x 2 + 4x m 2 x = 3 What would you like to do now?

Comments Begin Solution Question 3B Quadratics Menu Back to Home Continue Solution (a) 12m 7mhouse xmxm ?m?m garage the length of the garage should be 4m longer than its width and its floor area should be 25% of the floor area of the house. Length of garage = (x+4) m So area = x(x+4) = x 2 + 4x m 2 1. Area = length x width. Width is x but still need to find length.

Comments Begin Solution Question 3B Quadratics Menu Back to Home Continue Solution 2. Area = length x width. Find the required area and use your answer to (a) to establish quadratic equation. (b) 12m 7mhouse xmxm ?m?m garage the length of the garage should be 4m longer than its width and its floor area should be 25% of the floor area of the house. Area house = 12 x 7 = 84m 2 Area garage = 25% of 84m 2 = 21m 2 So x 2 + 4x = 21 x 2 + 4x - 21 = 0 (x + 7)(x – 3) = 0 (x + 7) = 0 or (x – 3) = 0 3. Make the equation equal to zero, factorise and solve.

Comments Begin Solution Question 3B Quadratics Menu Back to Home Continue Solution (b) 12m 7mhouse xmxm ?m?m garage the length of the garage should be 4m longer than its width and its floor area should be 25% of the floor area of the house. x 2 + 4x - 21 = 0 (x + 7)(x – 3) = 0 (x + 7) = 0 or (x – 3) = 0 3. Make the equation equal to zero, factorise and solve. ie. x = -7 or x = 3 Length must be 3m as negative value not valid. What would you like to do now?

Comments Quadratics Menu Back to Home Next Comment The quadratic equation can also be solved by applying the quadratic formula: ax 2 + bx + c = 0 Refer to the Formula Sheet: x = -b ±  (b 2 – 4ac ) 2a Quadratic Formula: Finding first eases working. Bracket all negative numbers. b 2 – 4ac 2. Area = length x width. Find the required area and use your answer to (a) to establish quadratic equation. Area house = 12 x 7 = 84m 2 Area garage = 25% of 84m 2 = 21m 2 So x 2 + 4x = 21 x 2 + 4x - 21 = 0 (x + 7)(x – 3) = 0 (x + 7) = 0 or (x – 3) = 0 3. Make the equation equal to zero, factorise and solve.

Comments Quadratics Menu Back to Home Next Comment Take care when allocating a value to a, b and c. ax 2 + bx + c = 0 1x 2 + 4x - 21 = 0 a =1 b = 4 c = (-21) 2. Area = length x width. Find the required area and use your answer to (a) to establish quadratic equation. Area house = 12 x 7 = 84m 2 Area garage = 25% of 84m 2 = 21m 2 So x 2 + 4x = 21 x 2 + 4x - 21 = 0 (x + 7)(x – 3) = 0 (x + 7) = 0 or (x – 3) = 0 3. Make the equation equal to zero, factorise and solve.

Comments Quadratics Menu Back to Home Next Comment Make sure that you understand when a negative answer may not be correct in context given, e.g. negative lengths or areas. (b) x 2 + 4x - 21 = 0 (x + 7)(x – 3) = 0 (x + 7) = 0 or (x – 3) = 0 3. Make the equation equal to zero, factorise and solve. ie. x = -7 or x = 3 Length must be 3m as negative value not valid.

(a)State the coordinates of E. (b)The second parabola cuts the X-axis at C & D. D is the point (8,0), find the coordinates of C. (c)Find the equation of the parabola with minimum turning point G. Quadratic Functions: Question 4 Go to full solution Go to Comments Reveal answer EXIT Get hint The blades in electric hair-clippers have grooves in the form of identical adjacent parabolas. The second parabola has turning point E and equation y = (x – 5) 2 – 9. X Y EFG CD

(a)State the coordinates of E. (b)The second parabola cuts the X-axis at C & D. D is the point (8,0), find the coordinates of C. (c)Find the equation of the parabola with minimum turning point G. In ‘completed square’, a minimum value occurs when bracket is zero. Quadratic Functions: Question 4 EXIT Use symmetry of a parabola to find other root if you know one. Go to full solution Go to Comments Go to Quadratics Menu Reveal answer The blades in electric hair-clippers have grooves in the form of identical adjacent parabolas. X Y EFG CD roots are equidistant from axis of symmetry. use ‘completed square’ form to write equation of parabola. What would you like to do now?

(a)State the coordinates of E. (b)The second parabola cuts the X-axis at C & D. D is the point (8,0), find the coordinates of C. (c)Find the equation of the parabola with minimum turning point G. The blades in electric hair-clippers have grooves in the form of identical adjacent parabolas. The second parabola has turning point E and equation y = (x – 5) 2 – 9. X Y EFG CD Quadratic Functions: Question 4 EXIT Try another like this Go to full solution Go to Comments Go to Quadratics Menu E is the point (5,-9). C is (2,0). y = (x – 17) 2 - 9

Comments Begin Solution Question 4 Quadratics Menu Back to Home Continue Solution 1. In ‘completed square’, a minimum value occurs when bracket is zero. X Y EFG CD The second parabola has turning point E and equation y = (x – 5) 2 – 9. (a) y will have a minimum value of -9 when (x – 5) 2 = 0. ie when x = 5 so E is the point (5,-9).

Comments Begin Solution Question 4 Quadratics Menu Back to Home Continue Solution 2. Use symmetry of a parabola to find other root if you know one. X Y EFG CD E is the point (5,-9). (b) D is the point (8,0), find the coordinates of C. (b)Horizontal dist from E to D = horizontal dist from E to C = 3 units D has x-coordinate 8 so x-coordinate of C is 6 units less ie C is (2,0).

Comments Begin Solution Question 4 Quadratics Menu Back to Home 3. Use symmetry of a parabola to find other turning points if you know one. X Y EFG CD E is the point (5,-9). (c)Distance from E to G is 12 units so G is (17,-9) 4. If you know turning point, use ‘completed square’ form to write equation of parabola. Equation of this parabola in same form as the second iey = (x – 17) 2 - 9 Try another like this (c) Find the equation of the parabola with minimum G. What would you like to do now?

Comments Quadratics Menu Back to Home Next Comment Learn the following Results: When a quadratic equation is in the form y = (x – a) 2 + b the parabola has a minimum T.P. at (a,b). When a quadratic equation is in the form y = b - (x – a) 2 the parabola has a maximum T.P. at (a,b). 1. In ‘completed square’, a minimum value occurs when bracket is zero. (a) y will have a minimum value of -9 when (x – 5) 2 = 0. ie when x = 5 so E is the point (5,-9).

Comments Quadratics Menu Back to Home Next Comment Learn the following Results: y = (x – 5) 2 - 9 Minimum T.P. at P(5,-9) e.g. P 1. In ‘completed square’, a minimum value occurs when bracket is zero. (a) y will have a minimum value of -9 when (x – 5) 2 = 0. ie when x = 5 so E is the point (5,-9).

Comments Quadratics Menu Back to Home Try another like this Learn the following Results: P The roots are equidistant from the turning point (or axis of symmetry) in the horizontal direction. 2. Use symmetry of a parabola to find other root if you know one. (b)Horizontal dist from E to D = horizontal dist from E to C = 3 units D has x-coordinate 8 so y-coordinate of C is 6 units less ie C is (2,0).

Quadratic Functions: Question 4B Go to full solution Go to Comments Reveal answer EXIT Get hint The top of an ornate fence consists of a series of congruent adjacent parabolas. The first parabola has turning point U and equation y = 4 – (x – 3) 2. (a)State the coordinates of U. (b)If T is the point (5,0) then find the coordinates of V, the maximum turning point on the second parabola. (c)Find the equation of the parabola with maximum turning point W. UVW X Y T

The top of an ornate fence consists of a series of adjacent parabolas. The first parabola has turning point U and equation y = 4 – (x – 3) 2. (a)State the coordinates of U. (b)If T is the point (5,0) then find the coordinates of V, the maximum turning point on the second parabola. (c)Find the equation of the parabola with maximum turning point W. UVW X Y T Quadratic Functions: Question 4B EXIT Go to full solution Go to Comments Go to Quadratics Menu Reveal answer In ‘completed square’, a minimum value occurs when bracket is zero. Use symmetry of a parabola to find another tp if you know one. use ‘completed square’ form to write equation of parabola. What would you like to do now?

(a)State the coordinates of U. (b)If T is the point (5,0) then find the coordinates of V, the maximum turning point on the second parabola. (c)Find the equation of the parabola with maximum turning point W. Quadratic Functions: Question 4B EXIT Try another like this Go to full solution Go to Comments Go to Quadratics Menu U is the point (3,4). V is (7,4). y = 4 – (x – 11) 2 The top of an ornate fence consists of a series of adjacent parabolas. The first parabola has turning point U and equation y = 4 – (x – 3) 2. UVW X Y T

Comments Begin Solution Question 4B Quadratics Menu Back to Home Continue Solution 1. In ‘completed square’, a maximum value occurs when bracket is zero. (a) y will have a maximum value of 4 when (x – 3) 2 = 0. ie when x = 3 so U is the point (3,4). The first parabola has turning point U and equation y = 4 – (x – 3) 2. UVW X Y T

Comments Begin Solution Question 4B Quadratics Menu Back to Home Continue Solution 2. Use symmetry of a parabola to find another turning point if you know one. (b) T is the point (5,0), find the coordinates of V. (b)Horizontal dist from U to T = horizontal dist from T to V = 2 units T has x-coordinate 5 so x-coordinate of V is 2 units more ie V is (7,4). UVW X Y T U is the point (3,4). For identical parabolae,maximum values same so y-coordinate = 4

Comments Begin Solution Question 4B Quadratics Menu Back to Home 3. Use symmetry of a parabola to find other turning points if you know one. (c)Distance from U to W is 8 units so W is (11,4) 4. If you know turning point, use ‘completed square’ form to write equation of parabola. Equation of this parabola in same form as the first iey = 4 - (x – 11) 2 Try another like this (c) Find the equation of the parabola with maximum W. UVW X Y T U is the point (3,4). What would you like to do now?

Comments Quadratics Menu Back to Home Next Comment Learn the following Results: When a quadratic equation is in the form y = (x – a) 2 + b the parabola has a minimum T.P. at (a,b). When a quadratic equation is in the form y = b - (x – a) 2 the parabola has a maximum T.P. at (a,b). 1. In ‘completed square’, a maximum value occurs when bracket is zero. (a) y will have a maximum value of 4 when (x – 3) 2 = 0. ie when x = 3 so U is the point (3,4).

Comments Quadratics Menu Back to Home Learn the following Results: y = 4 – (x – 3) 2 Maximum T.P. at P(3,4) e.g. P 1. In ‘completed square’, a maximum value occurs when bracket is zero. (a) y will have a maximum value of 4 when (x – 3) 2 = 0. ie when x = 3 so U is the point (3,4). Try another like this

(a)State the coordinates of G. (b)Find the coordinates of H and J. (c)Find the equation of the parabola with minimum turning point at F. Quadratic Functions: Question 4C Go to full solution Go to Comments Reveal answer EXIT Get hint The diagram below shows some identical adjacent parabolas. The last parabola has a minimum turning point at G and cuts the x-axis at H and J. Its equation is y = (x – 12) 2 – 4. X Y FG HJ

(a)State the coordinates of G. (b)Find the coordinates of H and J. (c)Find the equation of the parabola with minimum turning point at F. The last parabola has a minimum turning point at G and cuts the x-axis at H and J. Its equation is y = (x – 12) 2 – 4. In ‘completed square’, a minimum value occurs when bracket is zero. Quadratic Functions: Question 4C EXIT If not given a root you must solve the quadratic. Go to full solution Go to Comments Go to Quadratics Menu Reveal answer use ‘completed square’ form to write equation of parabola. The diagram below shows some identical adjacent parabolas. X Y FG HJ What would you like to do now?

(a)State the coordinates of G. (b)Find the coordinates of H and J. (c)Find the equation of the parabola with minimum turning point at F. Quadratic Functions: Question 4C EXIT Go to full solution Go to Comments Go to Quadratics Menu G is the point (12,-4). y = (x – 8) 2 - 4 The diagram below shows some identical adjacent parabolas. The last parabola has a minimum turning point at G and cuts the x-axis at H and J. Its equation is y = (x – 12) 2 – 4. X Y FG HJ H is (10,0) and J is (14,0) What would you like to do now?

Comments Begin Solution Question 4C Quadratics Menu Back to Home Continue Solution 1. In ‘completed square’, a minimum value occurs when bracket is zero. (a) y will have a minimum value of -4 when (x – 12) 2 = 0. ie when x = 12 so G is the point (12,-4). X Y FG HJ Its equation is y = (x – 12) 2 – 4.

Comments Begin Solution Question 4C Quadratics Menu Back to Home Continue Solution 2. If you are not given one of the roots you must solve quadratic to find roots. G is the point (12,-4). X Y FG HJ (b) Find the coordinates of H and J. (b)At H & J (x – 12) 2 – 4 = 0 so (x – 12) 2 = 4 ie x – 12 = -2 or 2 ie x = 10 or 14 H is (10,0) and J is (14,0) From diagram: y = (x – 12) 2 – 4.

X Y FG HJ Comments Begin Solution Question 4C Quadratics Menu Back to Home 3. Use symmetry of a parabola to find other turning points if you know one. (c)Distance from G to H is 2 units so F is 4 units horizontally from G. 4. If you know turning point, use ‘completed square’ form to write equation of parabola. Equation of this parabola in same form as the last (c) Find the equation of the parabola with minimum F. G is the point (12,-4). 1014 So F is the point (8,-4). iey = (x – 8) 2 - 4 Continue Solution What would you like to do now?

Comments Quadratics Menu Back to Home Next Comment Learn the following Results: When a quadratic equation is in the form y = (x – a) 2 + b the parabola has a minimum T.P. at (a,b). When a quadratic equation is in the form y = b - (x – a) 2 the parabola has a maximum T.P. at (a,b). 1. In ‘completed square’, a minimum value occurs when bracket is zero. (a) y will have a minimum value of -4 when (x – 12) 2 = 0. ie when x = 12 so G is the point (12,-4).

Comments Quadratics Menu Back to Home Next Comment Learn the following Results: y = (x – 12) 2 – 4 Minimum T.P. at P(12,-4) e.g. P 1. In ‘completed square’, a minimum value occurs when bracket is zero. (a) y will have a minimum value of -4 when (x – 12) 2 = 0. ie when x = 12 so G is the point (12,-4). End of Unit 3

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Further Trig Algebraic Operations Quadratic Functions EXIT.

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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Further Trig Algebraic Operations Quadratic Functions EXIT.

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Presentation on theme: "INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 3 : Further Trig Algebraic Operations Quadratic Functions EXIT."— Presentation transcript:

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