Question

If $x=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$ and $y=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$, find $x^2+xy+y^2$

• A. $4{\left(\frac{a+b}{a-b}\right)}^2-1.$
• B. $4{\left(\frac{a+b}{a-b}\right)}^2+1.$
• C. $4{\left(\frac{a-b}{a+b}\right)}^2-1.$
• D. $4{\left(\frac{a-b}{a+b}\right)}^2+1.$

Answer 1

The correct option is A $4{\left(\frac{a+b}{a-b}\right)}^2-1.$

$x=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$ $;$ $y=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$

$\Rightarrow x=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\times \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{a+b+2\sqrt{ab}}{a-b}=\frac{{(\sqrt{a}+\sqrt{b})}^2}{a-b}$

$\Rightarrow y=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\times \frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{a+b-2\sqrt{ab}}{a-b}=\frac{{(\sqrt{a}-\sqrt{b})}^2}{a-b}$

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

$\Rightarrow x+y=\frac{a+b+2\sqrt{ab}+a+b-2\sqrt{ab}}{a-b}=\frac{2(a+b)}{(a-b)}$

$\Rightarrow {(x+y)}^2=\frac{{4(a+b)}^2}{{(a-b)}^2}$

$\Rightarrow xy=\left(\frac{a+b+2\sqrt{ab}}{a-b}\right)\left(\frac{a+b-2\sqrt{ab}}{a-b}\right)$

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

$=\frac{a^2+b^2+2ab-4ab}{{(a-b)}^2}$

$=\frac{a+b^2-2ab}{{(a-b)}^2}$

$=\frac{{(a-b)}^2}{{(a-b)}^2}$

$=1$

$\therefore x^2+xy+y^2=4{\left(\frac{a+b}{a-b}\right)}^2-1.$

Hence, the answer is $4{\left(\frac{a+b}{a-b}\right)}^2-1.$