Question

If x=r,then the solution set of the equation root x+1+root x-1=1 is:

Answer 1

Question:-

If x = r, then the solution set of the equation

$\sqrt{x+1} + \sqrt{x-1} = 1$

${\blue{\underline{\underline{\bold{Solution:-}}}}}$

$\sqrt{x+1} + \sqrt{x-1} = 1$

$\implies \sqrt{x+1} = 1 - \sqrt{x-1}$

Squaring on both sides

$\implies {\sqrt{x+1}}^{2} = {1 + \sqrt{x-1}}^{2}$

$\implies x + 1 = 1 + x -1 - 2 \sqrt{x-1}</p><p></p><p>[tex]\implies 1 = -2(\sqrt{x-1})$

Squaring on both sides

$\implies 1 = 4 (x-1)$

$\implies x - 1 = \frac{1}{4}$

$\implies x = \frac {1}{4} + 1$

$\implies x = \frac{5}{4}$

Answer 2

Answer:

$If x = r, then the solution set of the equation\sqrt{x+1} + \sqrt{x-1} = 1 {\blue{\underline{\underline{\bold{Solution:-}}}}}\sqrt{x+1} + \sqrt{x-1} = 1 \implies \sqrt{x+1} = 1 - \sqrt{x-1}Squaring on both sides\implies {\sqrt{x+1}}^{2} = {1 + \sqrt{x-1}}^{2}\implies x + 1 = 1 + x -1 - 2 \sqrt{x-1}</p><p></p><p>[tex]\implies 1 = -2(\sqrt{x-1}) Squaring on both sides\implies 1 = 4 (x-1) \implies x - 1 = \frac{1}{4}\implies x = \frac {1}{4} + 1\implies x = \frac{5}{4}$

HOPE YOU LIKE IT