$(x+1) ^\frac{1}{2} - (x-1) ^\frac{1}{2} = (4x-1) ^\frac{1}{2}$ , then what is the value of x.

Given: $(x+1)^\frac{1}{2} - (x-1)^\frac{1}{2} = (4x-1)^\frac{1}{2}$.

To do: To find the value of $x$.

Solution:
 

$(x+1)^\frac{1}{2} - (x-1)^\frac{1}{2} = (4x-1)^\frac{1}{2}$

$\Rightarrow ( ( x+1)^\frac{1}{2} - ( x-1)^\frac{1}{2})^2 = ( ( 4x-1)^\frac{1}{2})^2$     [on squaring both sides]

$\Rightarrow ( (x+1)^\frac{1}{2})^2 -2(x+1)^\frac{1}{2}(x-1)^\frac{1}{2}+( (x-1)^\frac{1}{2})^2 = ( (4x-1)^\frac{1}{2})^2$

$\Rightarrow x+1-2\sqrt{ x+1}\sqrt{ x-1}+x-1=4x-1$

$\Rightarrow -2\sqrt{ x+1}\sqrt{ x-1}=4x-1-2x$

$\Rightarrow -2\sqrt{ x+1}\sqrt{ x-1}=2x-1$

On squaring both sides again,

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

$\Rightarrow 4(x+1)(x-1)=4x^2-4x+1$

$\Rightarrow 4( x^2-1)=4x^2-4x+1$

$\Rightarrow 4x^2-4=4x^2-4x+1$

$\Rightarrow -4=-4x+1$

$\Rightarrow -4+4x-1=0$

$\Rightarrow 4x-5=0$

$\Rightarrow 4x=5$

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

Thus, $x=\frac{5}{4}$.

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Updated on: 10-Oct-2022

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