If x and y are positive integers such that  xy+x+y = 71,  x2y +xy2 =880, then x 2+y2 is equal to

# If x and y are positive integers such that  xy+x+y = 71,  x2y +xy2 =880, then x 2+y2 is equal to

1. A
2. B
3. C
4. D

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### Solution:

$\because \mathrm{xy}+\mathrm{x}+\mathrm{y}=71⇒\mathrm{xy}+\left(\mathrm{x}+\mathrm{y}\right)=71$

and

are the roots of the quadratic equation.

$\begin{array}{rl}{\mathrm{t}}^{2}-71\mathrm{t}+880& =0\\ ⇒\left(\mathrm{t}-55\right)\left(\mathrm{t}-16\right)& =0\\ ⇒\mathrm{t}& =55,16\\ \mathrm{x}+\mathrm{y}& =16\text{and}\mathrm{xy}=55\end{array}$

So,${\mathrm{x}}^{2}+{\mathrm{y}}^{2}=\left(\mathrm{x}+\mathrm{y}{\right)}^{2}-2\mathrm{xy}=\left(16{\right)}^{2}-110=146$

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