Question

Find two consecutive positive odd numbers such that the square of their sum exceeds the sum of their squares by 126.​

Answer 1

x

x+2

...

(x+x+2)^2=x^2+(x+2)^2+126

(2x+2)^2=x^2+x^2+4x+4+126

4x^2+8x+4=2x^2+4x+130

2x^2+4x-126=0

a=2,b=4,c=-126

b^2-4ac=16+1008

b^2-4ac=1024

sqrt( 1024 )= 32

x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29

x1=(-4+ 32)/4

x1=7

x2=(-4-32)/4

x2= -9

7 & 9