Exams Joint Entrance Examination Mathematics Quadratic Equations and Inequalities Question 99 Question 99 of 157 Medium MCQ
If x is a solution of the equation, \[\sqrt {2x + 1} \] \[ - \sqrt {2x - 1} = 1,\] \[\,\,\left( {x \ge {1 \over 2}} \right),\] then \[\sqrt {4{x^2} - 1} \] is equal to :
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Given,
\[\sqrt {2x + 1} - \sqrt {2x - 1} = 1\]
\[ \Rightarrow \] \[\sqrt {2x + 1} = 1 + \sqrt {2x - 1} \]
Squaring both sides, we get
2x + 1 = = = 1 + 2x - 1 + 2\[\sqrt {2x - 1} \]
\[ \Rightarrow \] 1 = = = 2\[\sqrt {2x - 1} \]
\[ \Rightarrow \] 1 = = = 4(2x - 1)
\[ \Rightarrow \] 8x - 4 = = = 1
\[ \Rightarrow \] x = = = \[{5 \over 8}\]
So, \[\sqrt {4{x^2} - 1} \]
\[ = \sqrt {4\left( {{{25} \over {64}}} \right) - 1} \]
\[ = \sqrt {{{36} \over {64}}} \]
\[ = {6 \over 8}\]
\[ = {3 \over 4}\] Answer the question to unlock