Question

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\sf{1. \ \sqrt{\dfrac{3x}{x+1}}+\sqrt{\dfrac{x+1}{\sqrt3}}=2,}$

$\sf{when \ x\neq0 \ and \ x\neq-1}$

Solution:

$\sf{\leadsto{\sqrt{\dfrac{3x}{x+1}}+\sqrt{\dfrac{x+1}{\sqrt3}}=2}}$

$\sf{Let \ \sqrt{\dfrac{3x}{x+1}}=m}$

$\sf{\therefore{m+\dfrac{1}{m}=2}}$

$\sf{Multiply \ throughout \ by \ m}$

$\sf{\therefore{m^{2}+1=2m}}$

$\sf{\therefore{m^{2}-2m+1=0}}$

$\sf{\therefore{m^{2}-m-m+1=0}}$

$\sf{\therefore{m(m-1)-1(m-1)=0}}$

$\sf{\therefore{(m-1)(m-1)=0}}$

$\sf{\therefore{m=1 \ or \ 1}}$

$\sf{The \ value \ of \ m=1}$

$\sf{But, \ \sqrt{\dfrac{3x}{x+1}}=m}$

$\sf{\therefore{\sqrt{\dfrac{3x}{x+1}}=1}}$

$\sf{On \ squaring \ both \ sides, \ we \ get}$

$\sf{\therefore{\dfrac{3x}{x+1}=1}}$

$\sf{\therefore{3x=x+1}}$

$\sf{\therefore{x=\dfrac{1}{2}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ x \ is \ \dfrac{1}{2}.}}}$

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$\sf{2. \ 2(x+\dfrac{1}{x})^{2}-7(x+\dfrac{1}{x})+5=0,}$

$\sf{when \ x\neq0}$

Solution:

$\sf{\leadsto{2(x+\dfrac{1}{x})^{2}-7(x+\dfrac{1}{x})+5=0}}$

$\sf{Let \ x+\dfrac{1}{x}=m}$

$\sf{\therefore{2m^{2}-7m+5=0}}$

$\sf{\therefore{2m^{2}-2m-5m+5=0}}$

$\sf{\therefore{2m(m-1)-5(m-1)=0}}$

$\sf{\therefore{(2m-5)(m-1)=0}}$

$\sf{\therefore{m=\dfrac{5}{2} \ or \ 1}}$

$\sf{when \ m=\dfrac{5}{2}}$

$\sf{x+\dfrac{1}{x}=\dfrac{5}{2}}$

$\sf{Multiply \ throughout \ by \ 2x}$

$\sf{\therefore{2x^{2}-5x+2=0}}$

$\sf{\therefore{2x^{2}-4x-x+2=0}}$

$\sf{\therefore{2x(x-2)-1(x-1)=0}}$

$\sf{\therefore{(2x-1)(x-2)=0}}$

$\sf{\therefore{x=\dfrac{1}{2} \ or \ 2}}$

$\sf{when \ m=1}$

$\sf{x+\dfrac{1}{x}=1}$

$\sf{Multiply \ throughout \ by \ x, \ we \ get}$

$\sf{x^{2}-x+1=0}$

$\sf{By \ quadratic \ formula, \ we \ get}$

$\sf{x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$

$\sf{\therefore{x=\dfrac{1\pm\sqrt{-3}}{2}}}$

$\sf{These \ roots \ are \ not \ real \ so \ not \ admissible.}$

$\sf\purple{\tt{\therefore{The \ values \ of \ x \ are \ \dfrac{1}{2} \ and \ 2.}}}$

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$\sf{3. \ (x^{2}+\dfrac{1}{x^{2}})-5(x+\dfrac{1}{x})+6=0,}$

$\sf{when \ x\neq0}$

Solution:

$\sf{\leadsto{(x^{2}+\dfrac{1}{x^{2}})-5(x+\dfrac{1}{x})+6=0}}$

$\sf{In \ a^{2}+b^{2}=(a+b)^{2}-2ab}$

$\sf{If \ a=x \ and \ b=\dfrac{1}{x}}$

$\sf{x^{2}+\dfrac{1}{x^{2}}=(x+\dfrac{1}{x})^{2}-2x\times\dfrac{1}{x}}$

$\sf{\therefore{x^{2}+\dfrac{1}{x^{2}}=(x+\dfrac{1}{x})^{2}-2}}$

$\sf{Substitute \ in \ given \ equation}$

$\sf{\therefore{(x+\dfrac{1}{x})^{2}-2-5(x+\dfrac{1}{x})+6=0}}$

$\sf{Substitute \ x+\dfrac{1}{x}=m}$

$\sf{\therefore{m^{2}-2-5m+6=0}}$

$\sf{\therefore{m^{2}-5m+4=0}}$

$\sf{\therefore{m^{2}-m-4m+4=0}}$

$\sf{\therefore{m(m-1)-4(m-1)=0}}$

$\sf{\therefore{(m-1)(m-4)=0}}$

$\sf{\therefore{m=1 \ or \ 4}}$

$\sf{when \ m=1}$

$\sf{x+\dfrac{1}{x}=1}$

$\sf{Multiply \ throughout \ by \ x, \ we \ get}$

$\sf{x^{2}-x+1=0}$

$\sf{By \ quadratic \ formula, \ we \ get}$

$\sf{x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$

$\sf{\therefore{x=\dfrac{1\pm\sqrt{-3}}{2}}}$

$\sf{These \ roots \ are \ not \ real \ so \ not \ admissible.}$

$\sf{when \ m=4,}$

$\sf{x+\dfrac{1}{x}=4}$

$\sf{Multiply \ throughout \ by \ x, \ we \ get}$

$\sf{x^{2}-4x+1=0}$

$\sf{By \ quadratic \ formula}$

$\sf{x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$

$\sf{\therefore{x=\dfrac{4\pm\sqrt{12}}{2}}}$

$\sf{\therefore{x=2\pm\sqrt3}}$

$\sf\purple{\tt{\therefore{The \ values \ of \ x \ are \ 2+\sqrt3}}}$

$\sf\purple{\tt{and \ 2-\sqrt3.}}$