Question

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

Answer 1

Hint: We know the identity $(a - b)(a + b) = {a^2} - {b^2}$, apply this identity and square both the given equations and simplify it to get the desired answer.

Complete step-by-step answer:
So we are provided with the equation
$\sqrt {2x + 1} - \sqrt {2x - 1} = 1$
Now let squaring both side, we get
${(a - b)^2} = {a^2} + {b^2} - 2ab$
$
   \Rightarrow {\left( {\sqrt {2x + 1} } \right)^2} + {\left( {\sqrt {2x - 1} } \right)^2} - 2\left( {\sqrt {2x + 1} } \right)\left( {\sqrt {2x - 1} } \right) = 1 \\
   \Rightarrow 2x + 1 + 2x - 1 - 2\sqrt {4{x^2} - 1} = 1 \\
   \Rightarrow 4x - 2\sqrt {4{x^2} - 1} = 1 \\
   \Rightarrow 4x - 1 = 2\sqrt {4{x^2} - 1} \\
   \Rightarrow \sqrt {4{x^2} - 1} = \dfrac{{4x - 1}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 2 \right) \\
 $
Now let us rearrange the given equation.
$\sqrt {2x + 1} = 1 + \sqrt {2x - 1} $
We can write it like this, now squaring both side
$
  {\sqrt {2x + 1} ^2} = {1^2} + {\sqrt {2x - 1} ^2} + 2\sqrt {2x - 1} \\
  2x + 1 = 1 + 2x - 1 + 2\sqrt {2x - 1} \\
  1 = 2\sqrt {2x - 1} \\
  \sqrt {2x - 1} = \dfrac{1}{2} \\
 $
We got $\sqrt {2x - 1} = \dfrac{1}{2}$
Now squaring both side, we will get
$
  2x - 1 = \dfrac{1}{4} \\
  2x = \dfrac{1}{4} + 1 \\
  2x = \dfrac{5}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \dfrac{5}{8} \\
 $
So finally we get the value of x i.e. $\dfrac{5}{8}$
Now we calculate that
$\sqrt {4{x^2} - 1} = \dfrac{{4x - 1}}{2}$now putting $x = \dfrac{5}{8}$in this we get,
$\sqrt {4{x^2} - 1} = \dfrac{{4 \times \dfrac{5}{8} - 1}}{2} = \dfrac{3}{4}$
So our answer is $\dfrac{3}{4}$

Note: Alternative method:
We got the value of $\sqrt {2x - 1} = \dfrac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 1 \right)$
And also we are given that $\sqrt {2x + 1} - \sqrt {2x - 1} = 1$. So putting $\sqrt {2x - 1} = \dfrac{1}{2}$in the equation. We will get-
$
  \sqrt {2x + 1} - \dfrac{1}{2} = 1 \\
  \sqrt {2x + 1} = \dfrac{1}{2} + 1 \\
  \sqrt {2x + 1} = \dfrac{3}{2}\,\,\,\,\,\,\,\,\,\, \to \left( 2 \right) \\
 $
Multiplying (1) and (2) we get
$
  \sqrt {2x + 1} \times \sqrt {2x - 1} = \dfrac{1}{2}\,\, \times \dfrac{3}{2}\,\, \\
  \sqrt {4{x^2} + 1} = \dfrac{3}{4} \\
 $
Hence our answer is $\dfrac{3}{4}$.