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6.9 USING Equations that Factor

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1 6.9 USING Equations that Factor

2 Problem Solving Guidelines
Phase 1: UNDERSTAND the problem What am I trying to find? What Data am I Given? Have I ever solved a Similar problem? Phase 2: Develop and carry out a PLAN What strategies might I use to solve the problem? How can I correctly carry out the strategies I selected? Phase 3: Find the ANSWER and CHECK Does the proposed solution check? What is the answer to the problem? Does the answer seem reasonable? Have I stated the answer clearly?

3 Example #1 The product of one more than a number and one less than a number is 8. One more than a number: x + 1 One less than a number: x – 1 Product of both IS 8 (x + 1)(x – 1) = 8 π‘₯ 2 βˆ’1=8 π‘₯ 2 βˆ’1 βˆ’8=0 π‘₯ 2 βˆ’9=0 (x – 3)(x + 3) = 0 X = 3 or x = -3 Note: They both check out, so they are both solutions.

4 Example #2 The square of a number minus twice the number is 48. Find the number. Let x = the Number. The square of a number: 𝒙 𝟐 Twice the Number: 2x 𝒙 𝟐 βˆ’πŸπ’™=πŸ’πŸ– 𝒙 𝟐 - 2x – 48 = 0 (x – 8)(x + 6) = 0 x = 8 or x = -6 Note: They both Check Out, so they are both solutions.

5 Example # 3 The area of the foresail on a 12-meter racing yacht is square meters. The sail’s height is 8.75 meters greater than its base. Find its base and height. Area = 1 2 Γ— base Γ— height Let h = the sail’s height And h – 8.75 = the length of the sail’s base 1 2 Γ— 𝐻 βˆ’8.75 ×𝐻=93.75 (H – 8.75)H = 𝐻 2 βˆ’8.75𝐻=187.5 𝐻 2 βˆ’8.75𝐻 βˆ’187.5=0 (H – 18.75)(H + 10) = 0 H – = 0 OR H + 10 = 0 H = H = -10

6 EXAMPLE #4 The Product of two consecutive Integers is 156. Find the Integers. First Integer = x Second Integer = x + 1 X(x+1) = 156 π‘₯ 2 +π‘₯=156 π‘₯ 2 +π‘₯ βˆ’156=0 (x – 12)(x + 13) = 0 X – 12 = 0 OR x + 13 = 0 X = 12 OR x = -13 Integers are 12 and 13 OR Integers are -12 and -13

7 classwork 1.The Product of Seven Less Than a Number and eight less than a number is The Product of one more than a number and one less than the number is One more than twice the square of a Number is The Width of a rectangular card is 2cm less than the length. The area is 15 π‘π‘š 2 . Find the length and width. 5. The product of two consecutive Integers is 462. Find the Integers.

8 1.The Product of Seven Less Than a Number and eight less than a number is 0.
Seven less than a Number: x – 7 Eight less than a number: x – 8 Product: Multiply the 2 Together (x – 7)(x – 8) = 0 x – 7 = 0 or x – 8 = 0 x = 7 x = 8

9 2. The Product of one more than a number and one less than the number is 24.
One more than a number: x + 1 One less than a number: x – 1 (x + 1)(x – 1) = 24 π‘₯ 2 βˆ’1=24 π‘₯ = 0 (x + 5)(x – 5) = 0 X + 5 = 0 or x – 5 = 0 X = -5 or x = 5

10 3. One more than twice the square of a Number is 73.
The square of a number: π‘₯ 2 Twice that: 2 π‘₯ 2 One more than THAT: 2 π‘₯ π‘₯ 2 +1=73 2 π‘₯ 2 βˆ’72=0 2( π‘₯ ) = 0 2(x + 6)(x – 6) = 0 2 β‰  0 x + 6= 0 or x – 6 = 0 x = -6 x = 6

11 4. The Width of a rectangular card is 2cm less than the length
4. The Width of a rectangular card is 2cm less than the length. The area is 15 π‘π‘š 2 . Find the length and width. A = lw A = 15 π‘π‘š 2 Width(w) = l – 2 L(l – 2) = 15 𝐿 2 βˆ’2𝐿=15 𝐿 2 βˆ’2𝐿 βˆ’15=0 (L + 3)(L – 5) = 0 L = -3 or l = 5 Which one makes more sense? Can you have a Negative length? L = 5cm, Therefore W= 5 – 2 = 3cm

12 5. The product of two consecutive Integers is 462. Find the Integers.
First integer: x Second Integer: x + 1 x(x+1) = 462 π‘₯ 2 + x = 462 π‘₯ 2 +π‘₯ βˆ’462=0 (x – 21)(x + 22) = 0 X = 21 or x = -22 So Either 21 and 22 OR -21 and -22


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