Asked by LP
Find three consecutive odd integers such that the square of the first increased by product of the other two is 224.
Answers
Answered by
Reiny
let the middle number be x
then the 1st is x-2
and the 3rd is x+2
(x-2)^2 + x(x+2) = 224
x^2 - 4x + 4 + x^2 + 2x = 224
2x^2 -2x - 220 = 0
x^2 - x - 110 = 0
(x-11)(x+10) = 0
x = 11 or x = -10, but x is to be odd
so
x = 11
and the three numbers are 9, 11, and 13
check:
9^2 + 11(13) = 224 , YEAH!
then the 1st is x-2
and the 3rd is x+2
(x-2)^2 + x(x+2) = 224
x^2 - 4x + 4 + x^2 + 2x = 224
2x^2 -2x - 220 = 0
x^2 - x - 110 = 0
(x-11)(x+10) = 0
x = 11 or x = -10, but x is to be odd
so
x = 11
and the three numbers are 9, 11, and 13
check:
9^2 + 11(13) = 224 , YEAH!
Answered by
gricelis
i got this so correct omg im so happy