The equation sqrt{x + 1} - sqrt{x - 1} = sqrt | vedclass

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(A) Given equation: $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$.Squaring both sides: $(x + 1) + (x - 1) - 2\sqrt{(x + 1)(x - 1)} = 4x - 1$.$2x - 2\s

Vedclass · Accepted Answer

No solution

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(A) Given equation: $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$.Squaring both sides: $(x + 1) + (x - 1) - 2\sqrt{(x + 1)(x - 1)} = 4x - 1$.$2x - 2\sqrt{x^2 - 1} = 4x - 1$.$-2\sqrt{x^2 - 1} = 2x - 1$.Squaring both sides again: $4(x^2 - 1) = (2x - 1)^2$.$4x^2 - 4 = 4x^2 - 4x + 1$.$-4 = -4x + 1$.$4x = 5 \implies x = 5/4$.Checking $x = 5/4$ in the original equation: $\sqrt{5/4 + 1} - \sqrt{5/4 - 1} = \sqrt{9/4} - \sqrt{1/4} = 3/2 - 1/2 = 1$.However,the $RHS$ is $\sqrt{4(5/4) - 1} = \sqrt{5 - 1} = \sqrt{4} = 2$.Since $1 
eq 2$,the value $x = 5/4$ is an extraneous solution.Therefore,the equation has no solution.

The equation $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ has

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