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

$If x = \frac{\sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} - \sqrt{a^{2} - b^{2}}} then show that$ $b^{2} x^{2} - 2 a^{2} x + b^{2} = 0 .$
	Answered by Nimnim111118 last updated on 27/Dec/23

	$W e h a v e \frac{x}{1} = \frac{\sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} - \sqrt{a^{2} - b^{2}}}$ $\Rightarrow \frac{x + 1}{x - 1} = \frac{2 \sqrt{a^{2} + b^{2}}}{2 \sqrt{a^{2} - b^{2}}} (c o m p o n e n d o & d i v i d e n d o)$ $\Rightarrow {(\frac{x + 1}{x - 1})}^{2} = \frac{a^{2} + b^{2}}{a^{2} - b^{2}} (b y s q u a r i n g b o t h s i d e s)$ $\Rightarrow \frac{x^{2} + 2 x + 1}{x^{2} - 2 x + 1} = \frac{a^{2} + b^{2}}{a^{2} - b^{2}}$ $\Rightarrow \frac{2 x^{2} + 2}{4 x} = \frac{2 a^{2}}{2 b^{2}} (c o m p o n e n d o & d i v i d e n d o)$ $\Rightarrow \frac{x^{2} + 1}{2 x} = \frac{a^{2}}{b^{2}}$ $\Rightarrow b^{2} x^{2} + b^{2} = 2 a^{2} x$ $\Rightarrow b^{2} x^{2} - 2 a^{2} x + b^{2} = 0$
	Answered by deleteduser1 last updated on 27/Dec/23

	$x = \frac{{(\sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}})}^{2} = 2 (a^{2} + \sqrt{a^{4} - b^{4}})}{{(\sqrt{a^{2} + b^{2}})}^{2} - {(\sqrt{a^{2} - b^{2}})}^{2} = 2 b^{2}}$ $\Rightarrow {(b^{2} x - a^{2})}^{2} = a^{4} - b^{4} \Rightarrow b^{4} x^{2} - 2 a^{2} b^{2} x + a^{4} = a^{4} - b^{4}$ $\Rightarrow b^{2} x^{2} - 2 a^{2} x + b^{2} = 0$
	Answered by Rasheed.Sindhi last updated on 28/Dec/23
	$If x = (((√(a^2 + b^2 )) + (√(a^2 − b^2 )))/( (√(a^2 + b^2 )) − (√(a^2 − b^2 )))) then show that b^2 x^2 − 2a^2 x + b^2 = 0 b^2 x^2 − 2a^2 x + b^2 = 0 ⇒b^2 (x+(1/x))−2a^2 =0 (We need x+(1/x)) Let A=(√(a^2 + b^2 )) + (√(a^2 − b^2 )) & A^(−) =(√(a^2 + b^2 )) − (√(a^2 − b^2 )) x=(A/A^− ) , x+(1/x)=(A/A^− )+(A^− /A)=((A^2 +(A^(−) )^2 )/(AA^(−) )) =(((A+A^(−) )^2 −2AA^(−) )/(AA^(−) )) { ((A+A^(−) =2(√(a^2 +b^2 )) )),((AA^(−) =((√(a^2 +b^2 )) )^2 −((√(a^2 −b^2 )) )^2 =2b^2 )) :} x+(1/x)=(((2(√(a^2 +b^2 )))^2 −2(2b^2 ))/(2b^2 ))=((2a^2 )/b^2 ) b^2 x^2 − 2a^2 x + b^2 = 0 ⇒b^2 (x+(1/x))−2a^2 =0 ⇒b^2 (((2a^2 )/b^2 ))−2a^2 =0 ⇒0=0 (proved)$
	$If x = \frac{\sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} - \sqrt{a^{2} - b^{2}}} then show that$ $b^{2} x^{2} - 2 a^{2} x + b^{2} = 0$ $b^{2} x^{2} - 2 a^{2} x + b^{2} = 0$ $\Rightarrow b^{2} (x + \frac{1}{x}) - 2 a^{2} = 0 (W e n e e d x + \frac{1}{x})$ $L e t A = \sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}} & \overset{―}{A} = \sqrt{a^{2} + b^{2}} - \sqrt{a^{2} - b^{2}}$ $x = \frac{A}{\bar{A}}, x + \frac{1}{x} = \frac{A}{\bar{A}} + \frac{\bar{A}}{A} = \frac{A^{2} + {(\overset{―}{A})}^{2}}{A \overset{―}{A}}$ $= \frac{{(A + \overset{―}{A})}^{2} - 2 A \overset{―}{A}}{A \overset{―}{A}}$ ${\begin{cases} A + \overset{―}{A} = 2 \sqrt{a^{2} + b^{2}} \\ A \overset{―}{A} = {(\sqrt{a^{2} + b^{2}})}^{2} - {(\sqrt{a^{2} - b^{2}})}^{2} = 2 b^{2} \end{cases}$ $x + \frac{1}{x} = \frac{{(2 \sqrt{a^{2} + b^{2}})}^{2} - 2 (2 b^{2})}{2 b^{2}} = \frac{2 a^{2}}{b^{2}}$ $b^{2} x^{2} - 2 a^{2} x + b^{2} = 0$ $\Rightarrow b^{2} (x + \frac{1}{x}) - 2 a^{2} = 0$ $\Rightarrow b^{2} (\frac{2 a^{2}}{b^{2}}) - 2 a^{2} = 0$ $\Rightarrow 0 = 0 (p r o v e d)$
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