Question

If $xy=b$ and $\frac{1}{x^2}+\frac{1}{y^2}=a$, then ${(x+y)}^2$ equals:

• A. ${(a+2b)}^2$
• B. $a^2+b^2$
• C. $b(ab+2)$
• D. $ab(b+2)$
• E. $\frac{1}{a}+2b$

Answer 1

The correct option is C $b(ab+2)$

$\frac{1}{x^2}+\frac{1}{y^2}=a;$ $xy=b$

$\Rightarrow {(\frac{1}{x}+\frac{1}{y})}^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}$

$\Rightarrow {(\frac{1}{x}+\frac{1}{y})}^2=a+\frac{2}{b}$

$\Rightarrow (\frac{1}{x}+\frac{1}{y})=\frac{x+y}{xy}$

$\Rightarrow {(\frac{1}{x}+\frac{1}{y})}^2={\left(\frac{x+y}{xy}\right)}^2$

$\Rightarrow \frac{{(x+y)}^2}{{\left(xy\right)}^2}=a+\frac{2}{b}$

$\Rightarrow {(x+y)}^2=(a+\frac{2}{b}){\left(ay\right)}^2$

$=(a+\frac{2}{b}){\left(b\right)}^2$

$\Rightarrow {(x+y)}^2=a{b}^2+2b=b(ab+2).$

Hence, the answer is $b(ab+2).$