Question

Solve the following equations:
$\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{17}{4}$, $x^2+y^2=706$

• A. $x=\pm 9,y=\pm 25$
• B. $x=\pm 25,y=\pm 9$
• C. $x=\pm 36,y=\pm 49$
• D. $x=\pm 9,y=\pm 36$

Answer 1

The correct option is B $x=\pm 25,y=\pm 9$

Given equations are, $x^2+y^2=706$ .........(i)

and $\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{17}{4}$ .....(ii)

$\frac{{(\sqrt{x}-\sqrt{y})}^2+{(\sqrt{x}+\sqrt{y})}^2}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}=\frac{17}{4}\Rightarrow \frac{x+y-2\sqrt{x}\sqrt{y}+x+y+2\sqrt{x}\sqrt{y}}{x-y}=\frac{17}{4}\Rightarrow \frac{2x+2y}{x-y}=\frac{17}{4}\Rightarrow 8x+8y=17x-17y\Rightarrow 9x=25y\Rightarrow x=\frac{25}{9}y$

Substituting in (i), we get
${\left(\frac{25}{9}y\right)}^2+y^2=706\Rightarrow \frac{706}{81}{y}^2=706\Rightarrow y^2=81\Rightarrow y=\pm 9$

Substituting $y$ in (i), we get
$x^2+{(\pm 9)}^2=706\Rightarrow x^2=706-81\Rightarrow x^2=625\Rightarrow x=\pm 25$