Question

The only real value of $x$ satisfying the equation is $6\sqrt{\frac{x}{x+4}}-2\sqrt{\frac{x+4}{x}}=11$(1) where $x\in R$

• A. $4/35$
• B. $-4/35$
• C. $16/3$
• D. none of these.

Answer 1

Answer: (D)

For (1) to be defined either $x<-4$ or $x>0$.
Put $\sqrt{\frac{x}{x+4}}=y$. Equation (1) becomes
$6y-2/y=11\Rightarrow 6y^2-11y-2=0$
$\Rightarrow (6y+1)(y-2)=0\Rightarrow y=-1/6,y=2$
$\text{ As }y=\sqrt{\frac{x}{x+4}}\ge 0\text{, we reject }y=-\frac{1}{6}.$
$\text{ Thus, }y=2\Rightarrow \frac{x}{x+4}=y^2=4$
$\Rightarrow x=4x+16\Rightarrow x=-16/3<-4$