SSC CHSL 17 Mar 2020 Shift 3 Paper

Let the two numbers be $\mathit{x}$ and $\mathit{y}$.
According to the question,
$\mathit{x}+\mathit{y}=59$ ...(i)
and $\mathit{x}\times \mathit{y}=840$ ...(ii)
Now, on squaring Eq. (i) both sides we get,
${\mathit{x}}^2+{\mathit{y}}^2+2\mathit{x}\mathit{y}=3481$
${\mathit{x}}^2+{\mathit{y}}^2+2\times 840=3481$ [from Eq. (ii)]
⇒ ${\mathit{x}}^2+{\mathit{y}}^2+1680=3481$
⇒ ${\mathit{x}}^2+{\mathit{y}}^2=3481-1680$
⇒ ${\mathit{x}}^2+{\mathit{y}}^2=1801$
∴ Sum of squares $=1801$