Simplify √128 Simplify 4√45 Rationalise and simplify 2 5√2

8√2 12√5 √2 5

Quadratic graphs and equations Difference of two squares Solve the equation 𝑥 2 – 49 = 0 and hence sketch the graph 𝑦 = 𝑥 2 – 49. Chapter 3.1 𝑥 2 – 49 can be written as 𝑥 2 +0𝑥– 49 Notice that this is 𝑥 2 – 7 2   Quadratic Perfect square Difference of two squares 𝑥 2 +0𝑥– 49 = (𝑥 + 7)(𝑥 − 7) −7 + 7 = 0 and (−7) x 7 = −49  (𝑥 + 7)(𝑥 − 7) = 0 ⇒ 𝑥 = −7 or 𝑥 = 7

Quadratic graphs and equations Factorise 6 𝑥 2 + 𝑥 − 12. (ii) Solve the equation 6 𝑥 2 + 𝑥 − 12 = 0. (iii) Sketch the graph 𝑦 = 6 𝑥 2 + 𝑥 − 12. Chapter 3.1 Quadratic Perfect square Difference of two squares

Quadratic graphs and equations (𝑖) The technique for finding how to split the middle term is now adjusted. Start by multiplying the two outside numbers together: 6 × (−12) = −72. Now look for two numbers which add to give +1 (the coefficient of x) and multiply to give −72 (the number found above). (+9) + (−8) = +1 (+9) × (−8) = −72 Splitting the middle term gives Chapter 3.1 Quadratic Perfect square Difference of two squares 6 𝑥 2 + 9𝑥 − 8𝑥 − 12 = 3𝑥 (2𝑥 + 3) − 4(2𝑥 + 3) 3𝑥 is the largest factor of both 6 𝑥 2 and 9𝑥. −4 is the largest factor of both −8𝑥 and −12. = (3𝑥 − 4)(2𝑥 + 3)

Quadratic graphs and equations (𝑖𝑖) (3𝑥 − 4)(2𝑥 + 3) = 0 ⇒ 3𝑥 − 4 = 0 or 2𝑥 + 3 = 0 ⇒𝑥= 4 3 or 𝑥=− 3 2 (𝑖𝑖𝑖) The graph is a ∪-shaped curve which crosses the x axis at the points where 𝑥= 4 3 and where 𝑥=− 3 2 When 𝑥 = 0, 𝑦 = −12, so the curve crosses the y axis at (0, −12). Chapter 3.1 Quadratic Perfect square Difference of two squares

Quadratic graphs and equations (i) Factorise 15 + 2𝑥 − 𝑥 2 . (ii) Hence sketch the graph of 𝑦 = 15 + 2𝑥 − 𝑥 2 . Chapter 3.1 Quadratic Perfect square Difference of two squares

Quadratic graphs and equations The coefficient of 𝑥 2 is −1 so here the outside numbers are 15 and −1 (𝑖) Start by multiplying the two outside numbers together: 15 × −1 = −15 Now look for two numbers which add to give +2 (the coefficient of 𝑥) and multiply to give −15 (the number found above). (+5) + (−3) = +2 (+5) . (−3) = −15 Splitting the middle term gives Chapter 3.1 Quadratic Perfect square Difference of two squares 15+5𝑥−3 𝑥− 𝑥 2 = 5(3+𝑥) −𝑥(3+𝑥) =(5−𝑥)(3+𝑥)

Quadratic graphs and equations (ii) For numerically large positive or negative values of x, the value of 𝑦 = (5 − 𝑥)(3 + 𝑥) will be negative and consequently the curve will be ∩-shaped. It will cross the 𝑥 axis when 𝑦 = 0, i.e. at 𝑥 = +5 and 𝑥 = −3, and the 𝑦 axis when 𝑥 = 0, i.e. at 𝑦 = 15. (i) Factorise 15 + 2𝑥 − 𝑥 2 . (ii) Hence sketch the graph of 𝑦 = 15 + 2𝑥 − 𝑥 2 . Chapter 3.1 Quadratic Perfect square Difference of two squares

Quadratic graphs and equations Solve the quartic equation 4 𝑥 4 − 17 𝑥 2 + 4 = 0. Replacing 𝑥 2 by 𝑧 gives the quadratic equation 4 𝑧 2 − 17𝑧 + 4 = 0. Chapter 3.1 ⇒(𝑧 − 4)(4𝑧 − 1) = 0 Remember that you replaced 𝑥 2 by 𝑧 earlier. Quadratic Perfect square Difference of two squares ⇒𝑧 = 4 𝑜𝑟 4𝑧 = 1 ⇒ Either 𝑥 2 = 4 or 4 𝑥 2 = 1 ⇒The solution is 𝑥 =±2 or 𝑥=± 1 2

Quadratic graphs and equations In a right-angled triangle, the side BC is 1 cm longer than the side AB and the hypotenuse AC is 29 cm long. Find the lengths of the three sides of the triangle. Chapter 3.1 Quadratic Perfect square Difference of two squares

Quadratic graphs and equations Let the length of AB be x cm. So the length of BC is x + 1 cm. The hypotenuse is 29 cm. Using Pythagoras’ theorem: 𝑥 2 + 𝑥+1 2 = 29 2 Chapter 3.1 ⇒ 𝑥 2 + 𝑥 2 + 2𝑥 + 1 = 841 ⇒ 2 𝑥 2 + 2𝑥 − 840 = 0 Quadratic Perfect square Difference of two squares ⇒ 𝑥 2 + 𝑥 − 420 = 0 Divide by 2 to make the factorisation easier. ⇒ (𝑥 + 21)(𝑥 − 20) = 0 Notice that −21 is rejected since x is a length. ⇒ 𝑥 = −21 or 𝑥 = 20, giving 𝑥 + 1 = 21 The lengths of the three sides are therefore 20 cm, 21 cm and 29 cm. Make sure that you answer the question fully.

Select menu MODE A: Equation/Func Step 1: Select equation vs simultaneous equation solver. A polynomial is an expression of the form where the are constants, for example, or . The degree of a polynomial is the highest power, so the degree of a cubic equation, e.g. , is 3. So for quadratic equations, use 2. “Solve ” Choose polynomial mode, degree 3. Enter 1, -1, 3, -4, pressing = after each number. Use down arrows to scroll through solutions. You can use this to sketch graphs, solve equations, or check your factorisation if you are told not to use a calculator.

Solve the following either manually or with a calculator

The completed square form Chapter 3.2 The completed square form “It’s not that I’m so smart, it’s just that I stay with problems longer.” - Albert Einstein

The completed square form Write 𝑥 2 + 5𝑥 + 4 in completed square form. Hence state the equation of the line of symmetry and the coordinates of the vertex of the curve 𝑦 = 𝑥 2 + 5𝑥 + 4. 𝑥 2 + 5𝑥 + 4 5÷ 2= 5 2 , 5 2 2 = 25 4 Chapter 3.2 = 𝑥 2 + 5𝑥 + 25 4 + 4− 25 4 Parabola Vertex Completing the square = 𝑥+ 5 2 2 − 9 4 This is the completed square form. The line of symmetry is 𝑥+ 5 2 = 0 , or 𝑥=− 5 2 The vertex is (− 5 2 , − 9 4 ).

The completed square form Use the method of completing the square to write down the equation of the line of symmetry and the coordinates of the vertex of the curve 𝑦 = 2 𝑥 2 + 6𝑥 + 5. Chapter 3.2 Parabola Vertex Completing the square

The completed square form 2 𝑥 2 + 6𝑥 + 5=2 𝑥 2 + 3𝑥 + 5 2 First take out a factor of 2: Now proceed as before with the expression in the square brackets: =2 𝑥+ 3 2 2 − 3 2 2 + 5 2 Chapter 3.2 =2 𝑥+ 3 2 2 + 1 4 Parabola Vertex Completing the square =2 𝑥+ 3 2 2 + 1 2 The least value of the expression in the brackets is when 𝑥 + 3 2 = 0, so the equation of the line of symmetry is 𝑥= − 3 2 . The vertex is − 3 2 , 1 2 .

The completed square form Use the method of completing the square to find the equation of the line of symmetry and the coordinates of the vertex of the curve 𝑦 = − 𝑥 2 + 6𝑥 + 5. Chapter 3.2 Start by taking out a factor of (−1): − 𝑥 2 +6𝑥+ 5𝑥 = − 𝑥 2 −6𝑥 − 5 Completing the square for the expression in square brackets gives: Parabola Vertex Completing the square − 𝑥 2 −6𝑥 − 5 =− 𝑥−3 2 − 3 2 −5 = − 𝑥−3 2 −14 = − 𝑥−3 2 +14 The least value of the expression in the brackets is when 𝑥 − 3 = 0, so the equation of the line of symmetry is 𝑥 = 3. The vertex is (3, 14). This is a maximum point. It is a maximum since the coefficient of 𝑥 2 is negative.

The completed square form (i) Write the equation 𝑥 2 − 6𝑥 + 2 in completed square form. (ii) Hence solve the equation 𝑥 2 − 6𝑥 + 2 = 0. 𝑖 𝑥 2 − 6𝑥 + 2 = 𝑥 −3 2 − 3 2 +2 Chapter 3.2 = 𝑥− 3 2 − 7 𝑖𝑖 𝑥− 3 2 − 7= 0 Parabola Vertex Completing the square ⇒ 𝑥 −3 2 =7 Take the square root of both sides: ⇒𝑥−3= ± 7 ⇒𝑥 = 3 ± 7 This is an exact answer. Using your calculator to find the value of 7 ⇒𝑥 = 5.646 or 0.354, to 3 decimal places.  This is an approximate answer.

Select menu MODE A: Equation/Func “Solve ” Choose polynomial mode, degree 3. Enter 1, -1, 3, -4, pressing = after each number. Use down arrows to scroll through solutions. Step 1: Select equation vs simultaneous equation solver. A polynomial is an expression of the form where the are constants, for example, or . The degree of a polynomial is the highest power, so the degree of a cubic equation, e.g. , is 3. So for quadratic equations, use 2. You can use this to sketch the graph, minimum point, line of symmetry or check you have completed the square properly in a non-calculator question

Factorise, Solve, Complete the square and sketch each of the following labelling the roots, y-intercept, vertex and line of symmetry for each. (You can use your calculator to check your answers)

Simplify 3√2×5√2 Simplify√2( 6 +3 2 ) Simplify (𝑎+2 𝑏 )(𝑎−2 𝑏 )

30 6+2√3 a 2 −4 b 2

Chapter 3.3 The quadratic formula “It’s not that I’m so smart, it’s just that I stay with problems longer.” - Albert Einstein

The quadratic formula Use the quadratic formula to solve 3 𝑥 2 − 6𝑥 + 2 = 0. Chapter 3.3 Comparing it to the form 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐 = 0 gives 𝑎 = 3,𝑏 = −6,𝑐 = 2. 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Substituting these values into the formula  𝑥= −6± 36−24 6 gives = 0.423 or 1.577 (to 3 d.p.)

The quadratic formula Show that 𝑥 2 − 2𝑥 + 2 cannot be factorised. (ii) What happens when you try to use the quadratic formula to solve 𝑥 2 − 2𝑥 + 2=0? (iii) Draw the graph of 𝑦=𝑥 2 − 2𝑥 + 2. (iv) How does your graph explain your answer to (ii)? Chapter 3.3

The quadratic formula The only two whole numbers which multiply to give 2 are 2 and 1 (or−2 and −1) and they cannot be added to get −2. (ii) Comparing 𝑥 2 − 2𝑥 + 2 to the form 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐 = 0 gives 𝑎 = 1, 𝑏 = −2 and 𝑐 = 2. Chapter 3.3 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Substituting these values into the formula 𝑥= 2± 4−8 2 gives There is no real number equal to −4 so there is no solution. = 2± −4 2

The quadratic formula (iii) Chapter 3.3 (iv) Sketching the graph shows that it does not cut the x axis and so there is indeed no real solutions to the equation.

The Discriminant The discriminant is the part inside the square root b2-4ac It determines how many roots an equation will have and so how many times it crosses or touches the x axis If b2-4ac > 0 then the equation has two real roots If b2-4ac = 0 then the equation has one repeated root If b2-4ac < 0 then the equation has no real roots

How many roots does the equation 3x2 + 2x – 7 have?

For each equation below state the number of the roots of the equation and give solutions where possible in exact form

Solutions from the quadratic formula can be checked using your calculator