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If-x-a-2-b-2-a-2-b-2-a-2-b-2-a-2-b-2-then-show-that-b-2-x-2-2a-2-x-b-2-0-

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Question Number 202477 by MATHEMATICSAM last updated on 27/Dec/23
If x = (((√(a^2 + b^2 )) + (√(a^2 − b^2 )))/( (√(a^2 + b^2 )) − (√(a^2 − b^2 )))) then show that b^2 x^2 − 2a^2 x + b^2 = 0.
$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0}. \\ $$
Answered by Nimnim111118 last updated on 27/Dec/23
We have (x/1) = (((√(a^2 + b^2 )) + (√(a^2 − b^2 )))/( (√(a^2 + b^2 )) − (√(a^2 − b^2 )))) ⇒((x+1)/(x−1))=((2(√(a^2 +b^2 )))/(2(√(a^2 −b^2 )))) (componendo & dividendo) ⇒(((x+1)/(x−1)))^2 =((a^2 +b^2 )/(a^2 −b^2 )) (by squaring both sides) ⇒((x^2 +2x+1)/(x^2 −2x+1))=((a^2 +b^2 )/(a^2 −b^2 )) ⇒((2x^2 +2)/(4x))=((2a^2 )/(2b^2 )) (componendo & dividendo) ⇒((x^2 +1)/(2x))=(a^2 /b^2 ) ⇒b^2 x^2 +b^2 =2a^2 x ⇒b^2 x^2 −2a^2 x+b^2 =0
$${We}\:{have}\:\frac{{x}}{\mathrm{1}}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}=\frac{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:\left({componendo}\:\&\:{dividendo}\right) \\ $$$$\Rightarrow\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\left({by}\:{squaring}\:{both}\:{sides}\right) \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{4}{x}}=\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{2}{b}^{\mathrm{2}} }\:\left({componendo}\:\&\:{dividendo}\right) \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\: \\ $$$$\Rightarrow{b}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} {x} \\ $$$$\Rightarrow{b}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$
Answered by AST last updated on 27/Dec/23
x=((((√(a^2 +b^2 ))+(√(a^2 −b^2 )))^2 =2(a^2 +(√(a^4 −b^4 ))))/(((√(a^2 +b^2 )))^2 −((√(a^2 −b^2 )))^2 =2b^2 )) ⇒(b^2 x−a^2 )^2 =a^4 −b^4 ⇒b^4 x^2 −2a^2 b^2 x+a^4 =a^4 −b^4 ⇒b^2 x^2 −2a^2 x+b^2 =0
$${x}=\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{2}\left({a}^{\mathrm{2}} +\sqrt{{a}^{\mathrm{4}} −{b}^{\mathrm{4}} }\right)}{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} −\left(\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\left({b}^{\mathrm{2}} {x}−{a}^{\mathrm{2}} \right)^{\mathrm{2}} ={a}^{\mathrm{4}} −{b}^{\mathrm{4}} \Rightarrow{b}^{\mathrm{4}} {x}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {x}+{a}^{\mathrm{4}} ={a}^{\mathrm{4}} −{b}^{\mathrm{4}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$
Answered by Rasheed.Sindhi last updated on 28/Dec/23
If x = (((√(a^2 + b^2 )) + (√(a^2 − b^2 )))/( (√(a^2 + b^2 )) − (√(a^2 − b^2 )))) then show that b^2 x^2 − 2a^2 x + b^2 = 0 b^2 x^2 − 2a^2 x + b^2 = 0 ⇒b^2 (x+(1/x))−2a^2 =0 (We need x+(1/x)) Let A=(√(a^2 + b^2 )) + (√(a^2 − b^2 )) & A^(−) =(√(a^2 + b^2 )) − (√(a^2 − b^2 )) x=(A/A^− ) , x+(1/x)=(A/A^− )+(A^− /A)=((A^2 +(A^(−) )^2 )/(AA^(−) )) =(((A+A^(−) )^2 −2AA^(−) )/(AA^(−) )) { ((A+A^(−) =2(√(a^2 +b^2 )) )),((AA^(−) =((√(a^2 +b^2 )) )^2 −((√(a^2 −b^2 )) )^2 =2b^2 )) :} x+(1/x)=(((2(√(a^2 +b^2 )))^2 −2(2b^2 ))/(2b^2 ))=((2a^2 )/b^2 ) b^2 x^2 − 2a^2 x + b^2 = 0 ⇒b^2 (x+(1/x))−2a^2 =0 ⇒b^2 (((2a^2 )/b^2 ))−2a^2 =0 ⇒0=0 (proved)
$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\: \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{2}{a}^{\mathrm{2}} =\mathrm{0}\:\left({We}\:{need}\:{x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${Let}\:{A}=\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }\:\:\&\:\overline {{A}}=\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }\: \\ $$$${x}=\frac{{A}}{\overset{−} {{A}}}\:,\:{x}+\frac{\mathrm{1}}{{x}}=\frac{{A}}{\overset{−} {{A}}}+\frac{\overset{−} {{A}}}{{A}}=\frac{{A}^{\mathrm{2}} +\left(\overline {{A}}\right)^{\mathrm{2}} }{{A}\overline {{A}}} \\ $$$$\:\:\:=\frac{\left({A}+\overline {{A}}\right)^{\mathrm{2}} −\mathrm{2}{A}\overline {{A}}}{{A}\overline {{A}}} \\ $$$$\begin{cases}{{A}+\overline {{A}}=\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:}\\{{A}\overline {{A}}=\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\right)^{\mathrm{2}} −\left(\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\right)^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} }\end{cases} \\ $$$$\:{x}+\frac{\mathrm{1}}{{x}}=\frac{\left(\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{b}^{\mathrm{2}} \right)}{\mathrm{2}{b}^{\mathrm{2}} }=\frac{\mathrm{2}{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\:{b}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{2}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left(\frac{\mathrm{2}{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)−\mathrm{2}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}=\mathrm{0}\:\left({proved}\right) \\ $$

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