Question

Simplify (√a+x)+(√a-x)/(√a+x)-(√a-x) and find its value when x=2ab/1+b^2

Answer 1

Hey mate!

Here's your answer to a similar query!

If x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b)) … then prove that bx2-ax+b=0
x =(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))

use Component and dividend

(x+1)/(x-1) = √(a+2b)/√(a-2b)

square both sides

(x² + 2 x + 1)/(x^2-2x + 1) = (a+2b)/(a-2b)

again use component and dividend


(x² + 1)/(2x) = (a)/(2b)

or bx^2 +b = ax
or bx^2 - ax + b = 0

Answer 2

$= \frac{[ \sqrt{a}+x]+[ \sqrt{a}-x]}{[ \sqrt{a}+x]-[ \sqrt{a}-x]}$

$= \frac{[ \sqrt{a}+x]+[ \sqrt{a}-x]}{[ \sqrt{a}+x]-[ \sqrt{a}-x]} * \frac{[ \sqrt{a}+x]+[ \sqrt{a}-x]}{[ \sqrt{a}+x]+[ \sqrt{a}-x]}$

$= \frac{ [ \sqrt{a}+x]+[ \sqrt{a}-x]^{2} }{[ \sqrt{a}+x]^{2} -[ \sqrt{a}-x]^{2} }$

$= \frac{ [ \sqrt{a}+x]^{2}+[ \sqrt{a}-x]^{2}+2[ \sqrt{a}+x][ \sqrt{a}-x] }{[ \sqrt{a}+x]^{2} -[ \sqrt{a}-x]^{2} }$

$= \frac{4a}{4x \sqrt{a}}$

$= \frac{ \sqrt{a} }{x}$

Now if 
$x= \frac{2ab}{1+ b^{2} }$

then we get
$= \frac{ \sqrt{a} }{2ab}*[1+ b^{2}]$

$= \frac{ [1+ b^{2}] }{2b \sqrt{a} }$

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