Question

$\huge\red{\bold{\underline{\underline{Question:-}}}}$

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

Answer 1

TO PROVE :–

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$\bf \to{\sqrt{x+1}-\sqrt{x-1}\:=\:\sqrt{4x-1}}$ equation has no solution.

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SOLUTION :–

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

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• Square on both sides –

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

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• Using identity –

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$\bf \implies(a-b)^{2} \:= \: {a}^{2} + {b}^{2} - 2ab$

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

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

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

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• Using identity –

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$\bf \implies(a + b)(a-b)\:= \: {a}^{2} - {b}^{2}$

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

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

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

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• Again square on both sides –

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$\bf \implies (1 - 2x)^{2} \:=\:4({x}^{2} - 1)$

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

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$\bf \implies 1 - 4x \:=\:- 4$

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$\bf \implies 4x \:=\:5$

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$\bf \implies \large{ \boxed{ \bf x \:=\: \dfrac{5}{4}}}$

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✯ VERIFICATION :–

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$\bf \implies{\sqrt{ \dfrac{5}{4} +1}-\sqrt{ \dfrac{5}{4} -1}\:=\:\sqrt{4 \times \dfrac{5}{4} -1}}$

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

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

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$\bf \implies{ \dfrac{3}{2}-\dfrac{1}{2} \:=\:2}$

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$\bf \implies{ \dfrac{3 - 1}{2}\:=\:2}$

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$\bf \implies{ \dfrac{2}{2}\:=\:2}$

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$\bf \implies{1\:=\:2}$

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$\bf \implies x = \dfrac{5}{4} \: \: is \: \: not \: \: a \: \: solution \: \: of \: \: given \: \: equation.$

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$\bf \implies \:\:\: Hence\:\:Proved$

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▪︎ Hence , Given equation have no solution.

Answer 2

⟹√​x+1​​​−√​x−1​​​=√​4x−1​​​
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• Square on both sides –
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{( \sqrt{x+1}-\sqrt{x-1} )^{2} \:=:(\sqrt{4x-1}})^{2}⟹(√​x+1​​​−√​x−1​​​)​2​​=(√​4x−1​​​)​2​​
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• Using identity –
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(a-b)^{2} = {a}^{2} + {b}^{2} - 2ab⟹(a−b)​2​​=a​2​​+b​2​​−2ab
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( sqrt{x+1})^{2} + (sqrt{x-1} )^{2} - 2( \sqrt{x + 1})( sqrt{x - 1}) =4x-1⟹(√​x+1​​​)​2​​+(√​x−1​​​)​2​​−2(√​x+1​​​)(√​x−1​​​)=4x−1
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(x + 1)+ (x - 1)- 2( sqrt{x + 1})( sqrt{x - 1}) =4x-1⟹(x+1)+(x−1)−2(√​x+1​​​)(√​x−1​​​)=4x−1
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2x- 2\sqrt{(x + 1)(x - 1)}=:4x-1⟹2x−2√​(x+1)(x−1)​​​=4x−1
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• Using identity –
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(a + b)(a-b)= {a}^{2} - {b}^{2}⟹(a+b)(a−b)=a​2​​−b​2​​
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2x- 2\sqrt{ {x}^{2} - 1 }=:4x-1⟹2x−2√​x​2​​−1​​​=4x−1
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2x- 4x=2sqrt{ {x}^{2} - 1 }-1⟹2x−4x=2√​x​2​​−1​​​−1
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1 - 2x=2sqrt{ {x}^{2} - 1 }⟹1−2x=2√​x​2​​−1​​​
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• Again square on both sides –
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(1 - 2x)^{2} =4({x}^{2} - 1)⟹(1−2x)​2​​=4(x​2​​−1)
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1 + 4 {x}^{2} - 4x =4{x}^{2} - 4⟹1+4x​2​​−4x=4x​2​​−4
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1 - 4x = 4⟹1−4x=−4
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frac{5}{4}}}⟹​x=​4​​5​​​​​
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✯ VERIFICATION :–

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{\sqrt{ \dfrac{5}{4} +1}-\sqrt{ \dfrac{5}{4} -1}=:\sqrt{4 \times \dfrac{5}{4} -1}}⟹√​​4​​5​​+1​​​−√​​4​​5​​−1​​​=√​4×​4​​5​​−1​​​
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{\sqrt{ \dfrac{5 + 4}{4}}-\sqrt{ \dfrac{5 - 4}{4} }\:=\:\sqrt{5 -1}}⟹√​​4​​5+4​​​​​−√​​4​​5−4​​​​​=√​5−1​​​
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{\sqrt{ \dfrac{9}{4}}-\sqrt{ \dfrac{1}{4} }=:\sqrt{4}}⟹√​​4​​9​​​​​−√​​4​​1​​​​​=√​4​​​
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\bf \implies{ frac{3}{2}-frac{1}{2} =2}⟹​2​​3​​−​2​​1​​=2
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frac{3 - 1}{2}\:=\:2}⟹​2​​3−1​​=2
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{ frac{2}{2}=2}⟹​2​​2​​=2