Question

3 consecutive odd ontegers are such that the square of the third integer is 65 less than the sum of the squares of the first two. one solution is -7, -5, -3. find 3 other consecutive odd integers that also satisfy the given conditions

Answer 1

11, 13, 15

:))

Answer 2

OMG your like super smart cud u explain y

Answer 3

(x+4)^2=x^2+(x+2)^2-65

Answer 4

x^2-4x-77=0
(x-11)(x+7)=0
x=11 or x=-7

Answer 5

-7, -5, -3
11,13,15

Answer 6

our numbers are:

x -------->1st number
x+2 ----->2nd number
x+4-------> 3rd number

and our equation is:

x^2 + (x+2)^2 - 65 = (x+4)^2

then just simplify:
it will be:

x^2 - 4x -77 = 0

then get the value of x wich is 11, so:

x = 11
x+2 = 13
X+4 = 15

:))

Answer 7

omg thank you