Question

If x is a solution of the equation, $\sqrt{2x+1}-\sqrt{2x-1}=1,\left( x\ge \dfrac{1}{2} \right),\ \text{then }\sqrt{4{{x}^{2}}-1}$ is equal to
A) 2
B) $\dfrac{3}{4}$
C) $2\sqrt{2}$
D) $\dfrac{4}{7}$

Answer 1

Hint: For solving this problem, first we take one of the terms from left hand side to right hand side and then use the squaring of both sides to obtain the possible value of x. By doing so, we can easily evaluate the expression by replacing x.

Complete step-by-step answer:
According to the problem statement, we are given an expression $\sqrt{2x+1}-\sqrt{2x-1}=1,\left( x\ge \dfrac{1}{2} \right)$. To simplify this expression to obtain a value of x, we take $\sqrt{2x-1}$ term and shift it to the right side of the equation. On doing so, we get $\sqrt{2x+1}=1+\sqrt{2x-1}$.
Now, squaring both sides to simplify the expression by using the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we get ${{\left( \sqrt{2x+1} \right)}^{2}}={{\left( 1+\sqrt{2x-1} \right)}^{2}}$. As we know that the square of square root of a quantity is the same quantity itself, so
$\begin{align}
  & \Rightarrow 2x+1={{1}^{2}}+{{\left( \sqrt{2x-1} \right)}^{2}}+2\sqrt{2x-1} \\
 & \Rightarrow 2x+1=1+2x-1+2\sqrt{2x-1} \\
\end{align}$
On cancelling the common terms from both sides, we get
$\begin{align}
  & \Rightarrow 2\sqrt{2x-1}=1 \\
 & \Rightarrow \sqrt{2x-1}=\dfrac{1}{2} \\
\end{align}$
Now, squaring both sides, we get
$\begin{align}
  & \Rightarrow {{\left( \sqrt{2x-1} \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}} \\
 & \Rightarrow 2x-1=\dfrac{1}{4} \\
 & \Rightarrow 2x=1+\dfrac{1}{4} \\
 & \Rightarrow 2x=\dfrac{4+1}{4} \\
 & \Rightarrow x=\dfrac{5}{8} \\
\end{align}$
Since $\dfrac{5}{8}>\dfrac{1}{2}$, the stated conditions also hold true.
Now, finally replacing x in the desired expression, to obtain the simplified expression, we get
$\begin{align}
  & \Rightarrow \sqrt{4{{x}^{2}}-1}=\sqrt{4{{\left( \dfrac{5}{8} \right)}^{2}}-1} \\
 & \Rightarrow \sqrt{4\times \dfrac{25}{64}-1} \\
 & \Rightarrow \sqrt{\dfrac{25}{16}-1} \\
 & \Rightarrow \sqrt{\dfrac{25-16}{16}} \\
 & \Rightarrow \sqrt{\dfrac{9}{16}} \\
 & \Rightarrow \pm \dfrac{3}{4} \\
 & \therefore \dfrac{3}{4} \\
\end{align}$
Therefore, option (B) is correct.

Note: Students can re-verify that their obtained value of x is correct by using the condition given in the problem statement. On taking the square root of a quantity, we get both a positive and a negative result because the square of a positive and negative quantity is the same. By observing the options, we get option (B) as correct.