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)

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