 QUESTION

# If x is a solution of the equation $\sqrt{2x+1}-\sqrt{2x-1}=1,\left( x\ge \dfrac{1}{2} \right)$ , then $\sqrt{4{{x}^{2}}-1}$ is equal to(a) $\dfrac{3}{4}$ (b) $\dfrac{1}{2}$ (c) 2 (d) $2\sqrt{2}$

Hint: To solve the square root equation you need to follow the following steps: Isolate the square to one of the sides (L.H.S or R.H.S). Square both the sides of the given equation. Now solve the rest equation. By using the given expression, to find the value of x and then put the value of x in the required term.

Complete step-by-step solution -
The given expression is
$\sqrt{2x+1}-\sqrt{2x-1}=1$
$\sqrt{2x+1}=1+\sqrt{2x-1}$
Taking the squaring on both sides, we get
${{\left( \sqrt{2x+1} \right)}^{^{2}}}={{\left( 1+\sqrt{2x-1} \right)}^{2}}$
$2x+1={{\left( 1+\sqrt{2x-1} \right)}^{2}}$
Applying the formula ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ on left hand side, we get
$2x+1=1+2\left( \sqrt{2x-1} \right)+{{\left( \sqrt{2x-1} \right)}^{2}}$
$2x+1=1+2\left( \sqrt{2x-1} \right)+2x-1$
Cancelling the term $2x$ , we get
$1=2\left( \sqrt{2x-1} \right)$
Re-arranging the terms, we get
$2\left( \sqrt{2x-1} \right)=1$
Again, taking the squaring on both sides, we get
$4{{\left( \sqrt{2x-1} \right)}^{2}}=1$
We know that ${{\left( \sqrt{a} \right)}^{2}}=a$
$4\left( 2x-1 \right)=1$
Dividing both sides by 4, we get
$2x-1=\dfrac{1}{4}$
Adding one on both sides, we get
$2x=\dfrac{5}{4}$
Dividing both sides by 2, we get
$x=\dfrac{5}{8}$
This is the required value of the x which is to be put in the required expression.
Consider, the required expression
$\sqrt{4{{x}^{2}}-1}=\sqrt{4{{\left( \dfrac{5}{8} \right)}^{2}}-1}$
By using the formula ${{\left( \dfrac{a}{b} \right)}^{2}}=\dfrac{{{a}^{2}}}{{{b}^{2}}}$
$\sqrt{4{{x}^{2}}-1}=\sqrt{4\times\dfrac{25}{64}-1}=\sqrt{\dfrac{100}{64}-1}=\sqrt{\dfrac{100-64}{64}}$
$\sqrt{4{{x}^{2}}-1}=\sqrt{\dfrac{36}{64}}=\dfrac{6}{8}=\dfrac{3}{4}$
This is the required value of the term $\sqrt{4{{x}^{2}}-1}$
Therefore, the correct option for the given question is option (a).

Note: The possibility for the mistake is that you might get confused with the question that, Is the square Root of a Negative Number a whole number? The answer is no. As per the square root definition, negative numbers should not have a square root. Because if we multiply two negative numbers the result will always be a positive number. Square roots of negative numbers expressed as multiples of i (imaginary numbers).