Question

The sum of the squares of two consecutive multiples of 7 is 637 . Find the multiples .

Answer 1

Let the multiples of 7x&7(x+1)

$(7x{)}^2+(7(x+1){)}^2=637$

$\Rightarrow 49x^2+49(x+1{)}^2=637$

$\Rightarrow x^2+(x+1{)}^2=13$

$\Rightarrow x^2+x^2+1+2x-13=0$

$\Rightarrow 2x^2+2x-12=0$

$\Rightarrow x^2+x-6=0$

$\Rightarrow (x+3)(x-2)=0$

so x=-3 or x=2

no s= -14,-21 no s=14,21