If x is a solution of the equation sqrt{2x + | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(A) Given equation: $\sqrt{2x + 1} - \sqrt{2x - 1} = 1$  Squaring both sides: $(\sqrt{2x + 1} - \sqrt{2x - 1})^2 = 1^2$  $(2x + 1) + (2x - 1) - 2\sqrt{(2x + 1)(2x - 1)} = 1$  $4x - 2\sqrt{4x^2 - 1} = 1$  $4x - 1 = 2\sqrt{4x^2 - 1}$  Squaring again: $(4x - 1)^2 = 4(4x^2 - 1)$  $16x^2 - 8x + 1 = 16x^2 - 4$  $-8x = -5 \implies x = \frac{5}{8}$  Now,substitute $x = \frac{5}{8}$ into $\sqrt{4x^2 - 1}$:  $\sqrt{4(\frac{5}{8})^2 - 1} = \sqrt{4(\frac{25}{64}) - 1} = \sqrt{\frac{25}{16} - 1} = \sqrt{\frac{9}{16}} = \frac{3}{4}$

If $x$ is a solution of the equation $\sqrt{2x + 1} - \sqrt{2x - 1} = 1$ for $x \ge \frac{1}{2}$,then $\sqrt{4x^2 - 1}$ is equal to:

Similar Questions

What is the sum of all natural numbers $n$ such that the product of the digits of $n$ (in base $10$) is equal to $n^2-10n-36$?

Each of the roots of the equation $x^3-6x^2+6x-5=0$ are increased by $h$. If the new transformed equation does not contain the $x^2$ term,then $h$ is equal to:

The number of all common roots of the equation $x^4-10x^3+37x^2-60x+36=0$ and the transformed equation obtained by increasing any two distinct roots of it by $1$,keeping the other two roots fixed,is

If the roots of the equation $x^2 - 8x + a^2 - 6a = 0$ are real,then what is the range of $a$?

If $a$ and $b$ are natural numbers such that $2013 + a^2 = b^2$,then the minimum possible value of $ab$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module