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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

No solution

Vedclass · Answer

(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:

Similar Questions

Let $\alpha$ and $\beta$ be the roots of the equation $(x - a)(x - b) = c$,where $c \neq 0$. What are the roots of the equation $(x - \alpha)(x - \beta) + c = 0$?

Let $\theta$,$0 < \theta < \pi / 2$,be an angle such that the equation $x^2 + 4x \cos \theta + \cot \theta = 0$ has equal roots for $x$. Then $\theta$ in radians is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+3x^2-x-3=0$,then $(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = $

One of the real roots of the equation $x^3-6x^2+6x-2=0$ is

If the equation $(m - n)x^2 + (n - l)x + l - m = 0$ has equal roots,then $l, m,$ and $n$ satisfy:

Vedclass Test Series

Exam Paper Generator

Online Exam Module