Question

The number of solution(s) of the equation $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$ is/are

• A. $2$
• B. $0$
• C. $3$
• D. $1$

Answer 1

Answer: (B)

$0$

Given equation is
$\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$ .......(i)

On squaring both sides, we get
$(x+1)+(x-1)-2\sqrt{x^2-1}=4x-1$
$\Rightarrow 2x-2\sqrt{x^2-1}=4x-1$
$\Rightarrow -2\sqrt{x^2-1}=2x-1$

Again squaring both sides, we get
$4(x^2-1)=4x^2+1-4x$
$\Rightarrow -4=+1-4x$
$\Rightarrow 4x=5$
$\Rightarrow x=\frac{5}{4}$

But, when we put $x=\frac{5}{4}$ in equation (i), we get
$\sqrt{\frac{5}{4}+1}-\sqrt{\frac{5}{4}-1}=\sqrt{4\times \frac{5}{4}-1}$
$\Rightarrow \sqrt{\frac{9}{4}}-\sqrt{\frac{1}{4}}=\sqrt{5-1}\Rightarrow \frac{3}{2}-\frac{1}{2}=2$

$\Rightarrow 1=2$, which is not true.
Hence, no value of $x$ satisfy the given equation.