Question

Two squares have sides $x$ cm and $(x+4)$ cm. The sum of their areas is $656c{m}^2$.The value of x is:

Answer 1

$\Rightarrow$ Side of first square $=xcm$

$\Rightarrow$ Side of second square $=(x+4)cm$

According to the question,

$\Rightarrow$ $x^2+(x+4{)}^2=656$

$\Rightarrow$ $x^2+x^2+8x+16=656$

$\Rightarrow$ $2x^2+8x+16-656=0$

$\Rightarrow$ $2x^2+8x-640=0$

$\Rightarrow$ $x^2+4x-320=0$

$\Rightarrow$ $x^2+20x-16x-320=0$

$\Rightarrow$ $x(x+20)-16(x+20)=0$

$\Rightarrow$ $(x+20)(x-16)=0$

$\Rightarrow$ $x+20=0$ and $x-16=0$

$\Rightarrow$ $x=-20$ and $x=16$

Side cannot be negative.

$\therefore$ $x=16$