Question

If $\alpha$ satisfies the equation
$\sqrt{\frac{x}{2x+1}}+\sqrt{\frac{2x+1}{x}}=2$ , then the roots of the equation ${\alpha }^2{x}^2+4\alpha x+3=0$ are

• A. 1, 3
• B. -1, 1
• C. 2, -3
• D. 3, 4

Answer 1

Answer: (A)

$\sqrt{\frac{x}{2x+1}}+\sqrt{\frac{2x+1}{x}}=2\dots$(i)
Let $\sqrt{\frac{x}{2x+1}}=y$
Then, $y+\frac{1}{y}=2$
$\Rightarrow y^2-2y+1=0$
$\Rightarrow (y-1)(y-1)=0$
$\Rightarrow (y-1{)}^2=0$
$\Rightarrow y-1=0$
$\Rightarrow y=1$
$y=1$
$\Rightarrow \sqrt{\frac{x}{2x+1}}=1$
$\Rightarrow \frac{x}{2x+1}=1$
$\Rightarrow 2x+1=x$
$\Rightarrow x=-1$
$\because \alpha$ satisfies the Eq. (i),
therefore $\alpha =-1$
Now, ${\alpha }^2{x}^2+4\alpha x+3=0$
Put the value of $\alpha$
$(-1{)}^2{x}^2+4(-1)x+3=0$
$x^2-4x+3=0$
$(x-1)(x-3)=0$
$x=1,3$
Hence. the roots of the equation
${\alpha }^2{x}^2+4\alpha x+3=0$ are $1,3$