Question

Solve the following equations:
$x^{\frac{1}{2}}+y^{\frac{1}{2}}=5$,
$6(x^{-\frac{1}{2}}+y^{-\frac{1}{2}})=5$.

• A. $x=9;y=4$
• B. $x=4;y=9$
• C. $x=4;y=5$
• D. $x=5;y=4$

Answer 1

Answer: (A), (B)

The correct options are
A $x=9;y=4$
B $x=4;y=9$

Given, $\sqrt{x}+\sqrt{y}=5$ .....(i)

Also given, $\frac{6}{\sqrt{x}}+\frac{6}{\sqrt{y}}=5$

$\Rightarrow \frac{6(\sqrt{x}+\sqrt{y})}{\sqrt{xy}}=5\Rightarrow \frac{6\left(5\right)}{\sqrt{xy}}=5\Rightarrow \sqrt{xy}=6$ .......(ii)

$\Rightarrow \sqrt{y}=\frac{6}{\sqrt{x}}$

Substituting value of $\sqrt{y}$ in equation (i), we get

$\sqrt{x}+\frac{6}{\sqrt{x}}=5\Rightarrow \frac{x+6}{\sqrt{x}}=5\Rightarrow x+6=5\sqrt{x}\Rightarrow x^2+36+12x=25x\Rightarrow x^2-13x+36=0\Rightarrow x^2-9x-4x+36=0\Rightarrow x(x-9)-4(x-9)=0\Rightarrow (x-4)(x-9)=0\Rightarrow x=4,9$

Now from (ii), we have
$\sqrt{xy}=6\Rightarrow xy=36\Rightarrow y=\frac{36}{x}$

From $x=4$, we have

$y=\frac{36}{4}=9$

From $x=9$, we have

$y=\frac{36}{9}=4$