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**6.9 Using Equations that Factor**

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**1. The product of two consecutive positive even integers is 48**

1. The product of two consecutive positive even integers is 48. Find the two integers. Let n = first even integer (n + 2) = 2nd even integer n(n + 2) = 48 6 and 8 n2 + 2n = 48 n2 + 2n - 48= 0 (n + 8)(n – 6)= 0 n = -8, 6 X

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**2. The product of two consecutive positive odd integers is 143**

2. The product of two consecutive positive odd integers is 143. Find the two integers. Let n = first odd integer (n + 2) = 2nd odd integer n(n + 2) = 143 11 and 13 n2 + 2n = 143 n2 + 2n - 143= 0 (n - 11)(n + 13)= 0 n = -13, 11 X

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**X (x+2)(x +6) = 60 x2 + 8x + 12 = 60 x2 + 8x - 48= 0**

3. One side of a rectangle is 4 in. longer than the other. If the sides are each increased by 2 in., the area of the new rectangle is 60 in2. How long are the sides of the original rectangle? X X+4 (x+2)(x +6) = 60 x2 + 8x + 12 = 60 X+2 X+4+2 x2 + 8x - 48= 0 A = 60 in2 (x + 12)(x – 4)= 0 4 and 8 x = -12, 4 X

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**4. The sum of the squares of two consecutive even integers is 244**

4.The sum of the squares of two consecutive even integers is 244. Find the two integers. Let n² = first consecutive square Let (n+2)² = the 2nd consecutive square (n)² +(n+2) ²=244 n² +n² +4n +4 =244 2n² +4n +4 =244 2(n²+2n +2) =244 n² +2n +2 =122 n²+2n -120 =0 (n-10) (n+12) =0 n-10 =0 or n +12 =0 If n=10 then n+2 = 12 If n =-12 then n+2 =-10 10 and 12. Also, -12 and -10

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**5. Fifteen more than the square of a number is eight times the number.**

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**6. If seven is added to the square of a number, the result is 32**

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**7. The product of two consecutive integers is 182.**

13,14 and –13, -14

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**8. If you subtract a number from four times its square, the result is three.**

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**9.The sum of 6 times a positive number and 1 is the same as the square of 1 less than the number.**

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**Assignment: Page 294 (2-24) even**

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