• # question_answer Given that$\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$ Statement-1: Given equation has no solution Statement-2: $x=\frac{5}{4}$is an extraneous solution of given equation. A)  Statement -1 is true, statement -2 is true and statement-2 is correct explanation for statement -1. B)  Statement -1 is true, statement -2 is true and Statement-2 is NOT correct explanation for statement -1. C)  Statement-1 is true, Statement-2 is false D)  Statement-1 is false, statement -2 is true

Solution :

On squaring we get $x+1+x-1-2\sqrt{{{x}^{2}}-1}=4x-1$ $-2\sqrt{{{x}^{2}}-1}=2x-1$ On again squaring we get $4{{x}^{2}}-4=4{{x}^{2}}-4x+1$$\Rightarrow$$4x=5$$\Rightarrow$$x=\frac{5}{4}$ as if $x=\frac{5}{4}$- then second square root in given equation become imaginary, so no root is there.

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