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

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

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(D) Given equation: $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ $(i)$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 of $(i)$:$(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$Since $x \ge 1$,the left side $-2\sqrt{x^2 - 1}$ is $\le 0$,while the right side $2x - 1$ is $\ge 2(1) - 1 = 1$.Since a non-positive number cannot equal a positive number,there is no real value of $x$ that satisfies the equation.Thus,the equation has no solution.

The equation $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ for $x \in R$ has:

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