Question

Three consecutive odd integers are such that the square of the third integer is 9 less than the sum of the squares of the first two. One solution is −3, −1, and 1. Find three other consecutive odd integers that also satisfy the given conditions.

Answer 1

Let integers are x, x+2, x+4

(x+4)^2 = x^2 + (x+2)^2 - 9 the equation
x^2 + 8x + 16 = x^2 + x^2 + 4x + 4 - 9 subtracting the left side of equation

0= x^2 - 4x -21
factor this and test the values

Answer 2

do you need more help

Answer 3

(BTW - you know that x=-3 is one of the factors , given in the question)

Answer 4

Yes

Answer 5

Is the answer a decimal

Answer 6

It tells you that they are integers
@zpupster has given the correct equation - which is a quadratic:

0= x^2 - 4x -21

You are told that x=-3 is one solution

Find the other solution (by factorising)

if the other solution is m then the 3 numbers are m, m+2, m+4

Answer 7

can you factor this
??