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6.9 Using Equations that Factor

6.9 Using Equations that Factor. 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. n 2 + 2n = 48. n 2 + 2n - 48= 0. (n + 8)(n – 6)= 0. X. n = -8, 6.

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6.9 Using Equations that Factor

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  1. 6.9 Using Equations that Factor

  2. 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 X n = -8, 6

  3. 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 X n = -13, 11

  4. X+4 X X+4+2 X+2 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+2)(x +6) = 60 x2 + 8x + 12 = 60 x2 + 8x - 48= 0 A = 60 in2 (x + 12)(x – 4)= 0 4 and 8 X x = -12, 4

  5. 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

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

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

  8. 7. The product of two consecutive integers is 182. 13,14 and –13, -14

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

  10. Assignment:Page 294(2-20) even

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