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$\huge\red{\bold{\underline{\underline{Question:-}}}}$Prove that the equation, $\large\rm{\sqrt{x+1}-\sqrt{x-1}\:=\:\sqrt{4x-1}}$ has no solution​

Answer 1

Answer:

Given:-

$\sf \sqrt{x + 1} - \sqrt{x + 1} = \sqrt{4x - 1}$

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

$\sf \: - 2 \sqrt{x {}^{2} - 1 } \: \: \: 2x - 1$

$\sf \: 4x {}^{2} - 4 = 4 {}^{2} + 1 - 4x {}^{}$

$\sf \: 4x = - 5 = > x = - \frac{5}{4}$

$\sf \: But \: of \: x \: = - \frac{5}{4} \sqrt{x + 1} , \sqrt{x - 1} , \: \sqrt{4x - 1}$

Are not Define

Therefore,No solution

Answer 2

ANSWER:

To Prove:

• $\rm{\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}}$

Proof:

We are given that,

$\implies\rm{\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}- - - - (1)}$

Squaring both sides,

$\implies\rm{(\sqrt{x+1}-\sqrt{x-1})^2=(\sqrt{4x-1})^2}$

We know that,

$\hookrightarrow (a-b)^2=a^2+b^2-2ab$

So,

$\implies\rm{(\sqrt{x+1}-\sqrt{x-1})^2=(\sqrt{4x-1})^2}$

$\implies\rm{(\sqrt{x+1})^2+(\sqrt{x-1})^2-2(\sqrt{x+1})(\sqrt{x-1})=4x-1}$

$\implies\rm{x+1\!\!\!/+x-1\!\!\!/-2\sqrt{(x+1)(x-1)}=4x-1}$

We know that,

$\hookrightarrow (a+b)(a-b)=(a)^2-(b)^2$

So,

$\implies\rm{2x-2\sqrt{(x+1)(x-1)}=4x-1}$

$\implies\rm{2x-2\sqrt{(x)^2-(1)^2}=4x-1}$

$\implies\rm{2x-2\sqrt{x^2-1}=4x-1}$

Transposing 2x to RHS,

$\implies\rm{-2\sqrt{x^2-1}=4x-1-2x}$

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

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

Squaring both sides,

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

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

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

$\implies\rm{4x^2\!\!\!\!\!\!/\:-4=1+4x^2\!\!\!\!\!\!/\:-4x}$

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

On rearranging,

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

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

$\implies\rm{x=\dfrac{5}{4}}$

Now, we will substitute this in (1),

$\implies\rm{\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}}$

$\implies\rm{\sqrt{\dfrac{5}{4}+1}-\sqrt{\dfrac{5}{4}-1}=\sqrt{4\!\!\!/\left(\dfrac{5}{4\!\!\!/}\right)-1}}$

$\implies\rm{\sqrt{\dfrac{5+4}{4}}-\sqrt{\dfrac{5-4}{4}}=\sqrt{5-1}}$

$\implies\rm{\sqrt{\dfrac{9}{4}}-\sqrt{\dfrac{1}{4}}=\sqrt{4}}$

$\implies\rm{\dfrac{3}{2}-\dfrac{1}{2}=2}$

$\implies\rm{\dfrac{3-1}{2}=2}$

$\implies\rm{\dfrac{2}{2}=2}$

$\implies\rm{1=2}$

As, LHS ≠ RHS, the value of x, is not valid.

Hence no solution exists.

HENCE PROVED.