Question

If $\frac{9}{y}+\frac{4}{x}=\frac{12}{\sqrt{xy}}$ where $x>0,y>0$, then $3\sqrt{x}-2\sqrt{y}=$

• A. $5$
• B. $1$
• C. $2$
• D. $0$

Answer 1

The correct option is D $0$

$\frac{9}{y}+\frac{4}{x}=\frac{12}{\sqrt{xy}}$

${\left(\frac{3}{\sqrt{y}}\right)}^2+{\left(\frac{2}{\sqrt{x}}\right)}^2=\frac{12}{\sqrt{xy}}$

$\Rightarrow {(\frac{3}{\sqrt{y}}+\frac{2}{\sqrt{x}})}^2-2\frac{6}{\sqrt{xy}}=\frac{12}{\sqrt{xy}}$

$\Rightarrow {(\frac{3}{\sqrt{y}}+\frac{2}{\sqrt{x}})}^2-\frac{12}{\sqrt{xy}}=\frac{12}{\sqrt{xy}}$

$\Rightarrow {(\frac{3}{\sqrt{y}}+\frac{2}{\sqrt{x}})}^2=\frac{24}{\sqrt{xy}}$

$\Rightarrow {(3\sqrt{x}+2\sqrt{y})}^2=\frac{24}{\sqrt{xy}}xy$

$\Rightarrow {(3\sqrt{x}+2\sqrt{y})}^2=\frac{24}{\sqrt{xy}}xy\times \frac{\sqrt{xy}}{\sqrt{xy}}$

$\Rightarrow {(3\sqrt{x}+2\sqrt{y})}^2=\frac{24}{xy}xy\times \sqrt{xy}$

$\Rightarrow {(3\sqrt{x}+2\sqrt{y})}^2=24\sqrt{xy}$

$\Rightarrow {(3\sqrt{x}+2\sqrt{y})}^2-24\sqrt{xy}=0$

$\Rightarrow 9x+4y+12\sqrt{xy}-24\sqrt{xy}=0$

$\Rightarrow 9x+4y-12\sqrt{xy}=0$

$\therefore {(3\sqrt{x}-2\sqrt{y})}^2=0$

or $3\sqrt{x}-2\sqrt{y}=0$