The number of solution(s) of the equation sqr | vedclass

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(B) Given equation is $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$.For the square roots to be defined,we must have $x+1 \ge 0$,$x-1 \ge 0$,and $4x-1 \ge 0$,whi

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(B) Given equation is $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$.For the square roots to be defined,we must have $x+1 \ge 0$,$x-1 \ge 0$,and $4x-1 \ge 0$,which implies $x \ge 1$.Squaring both sides,we get $(x+1) + (x-1) - 2\sqrt{x^2-1} = 4x-1$.This simplifies to $2x - 2\sqrt{x^2-1} = 4x-1$,or $-2\sqrt{x^2-1} = 2x-1$.Squaring both sides again,we get $4(x^2-1) = (2x-1)^2 = 4x^2 - 4x + 1$.This simplifies to $4x^2 - 4 = 4x^2 - 4x + 1$,which gives $4x = 5$,or $x = \frac{5}{4}$.Checking $x = \frac{5}{4}$ in the original equation: $\sqrt{\frac{5}{4}+1} - \sqrt{\frac{5}{4}-1} = \sqrt{\frac{9}{4}} - \sqrt{\frac{1}{4}} = \frac{3}{2} - \frac{1}{2} = 1$.However,the right side is $\sqrt{4(\frac{5}{4})-1} = \sqrt{5-1} = \sqrt{4} = 2$.Since $1 
eq 2$,$x = \frac{5}{4}$ is an extraneous solution.Therefore,there are no real solutions to the equation.

The number of solution$(s)$ of the equation $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$ is/are

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