Question

Determine the values of 3 consecutive odd integers such that the sum of the squares of the first 2 is 15 less than the 3rd.

Answer 1

Let x be the integer.
x+2 = second odd integer
x+4 = third consec. odd integer

(x^2)+(x+2)^2= (x+4)-15

When it saaid 15 less than the 3rd, does that mean 15 less than the square of the third integer?

Answer 2

I believe so ..
I did the exact same thing ..
but i just dont understand what im doing :S

Answer 3

assume the first integer to be (2x+1), the second (2x+3), the third (2x+5)

Answer 4

\[x^{2}+(x+2)^{2}=(x+4)^{2}-15\]
\[2x^{2}+4x+4=x^{2}+8x+1\]
\[x^{2}+4x+3=0\]
\[(x+1)(x+3)=0\]
This is what I got .. but I need three numbers not 2 :S
And we have to solve them by using roots ..

Answer 5

the numbers are -1,-3 and -5

Answer 6

??

Answer 7

1,3 and 5 as well

Answer 8

1^2+3^2=5^2-15.. right?

Answer 9

im really confused ..
:S