Question

The value of $x$ satisfying the equation $6\sqrt{\frac{x}{x+4}}-2\sqrt{\frac{x+4}{x}}=11$ is

• A. $\frac{4}{35}$
• B. $-\frac{4}{35}$
• C. $\frac{16}{3}$
• D. None of these.

Answer 1

The correct option is D None of these.

$6\sqrt{\frac{x}{x+4}}-2\sqrt{\frac{x+4}{x}}=11$
For the square roots to exist
$\frac{x}{x+4}\ge 0\Rightarrow x<-4\text{or}x\ge 0\cdots \left(1\right)$
$\frac{x+4}{x}\ge 0\Rightarrow x\le -4\text{or}x>0\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$x\in (-\infty ,-4)\cup (0,\infty )$

Assuming $\sqrt{\frac{x}{x+4}}=y$
The given equation can be written as,
$6y-\frac{2}{y}=11\Rightarrow 6y^2-11y-2=0$
$\Rightarrow (6y+1)(y-2)=0\Rightarrow y=-\frac{1}{6},y=2\Rightarrow y=2(\because y=\sqrt{\frac{x}{x+4}}\ge 0)$
Thus,
$y=2\Rightarrow \frac{x}{x+4}=y^2=4\Rightarrow x=4x+16\Rightarrow x=-\frac{16}{3}$
This is a acceptable value of $x$ as $x\in (-\infty ,-4)\cup (0,\infty )$