Question:
Two squares have sides $x \mathrm{~cm}$ and $(x+4) \mathrm{cm}$. The sum of their areas is $656 \mathrm{~cm}^{2}$. Find the sides of the squares.
Solution:
Given that the sides of two square be $x \mathrm{~cm}$ and $(x+4) \mathrm{cm}$
Then according to question
$x^{2}+(x+4)^{2}=656$
$x^{2}+x^{2}+8 x+16=656$
$2 x^{2}+8 x+16-656=0$
$2 x^{2}+8 x-640=0$
$x^{2}+4 x-320=0$
$x^{2}+20 x-16 x-320=0$
$x(x+20)-16(x+20)=0$
$(x+20)(x-16)=0$
$(x+20)=0$
$x=-20$
Or
$(x-16)=0$
$x=16$
Since, sides of the squares being a positive, so x cannot be negative.
Therefore,
When $x=16$ then
$x+4=16+4$
$=20$
Thus, sides of the squares be $16 \mathrm{~cm}, 20 \mathrm{~cm}$