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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.
Ifx=a2+b2+a2−b2a2+b2−a2−b2thenshowthatb2x2−2a2x+b2=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
Wehavex1=a2+b2+a2−b2a2+b2−a2−b2⇒x+1x−1=2a2+b22a2−b2(componendo&dividendo)⇒(x+1x−1)2=a2+b2a2−b2(bysquaringbothsides)⇒x2+2x+1x2−2x+1=a2+b2a2−b2⇒2x2+24x=2a22b2(componendo&dividendo)⇒x2+12x=a2b2⇒b2x2+b2=2a2x⇒b2x2−2a2x+b2=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=(a2+b2+a2−b2)2=2(a2+a4−b4)(a2+b2)2−(a2−b2)2=2b2⇒(b2x−a2)2=a4−b4⇒b4x2−2a2b2x+a4=a4−b4⇒b2x2−2a2x+b2=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)
Ifx=a2+b2+a2−b2a2+b2−a2−b2thenshowthatb2x2−2a2x+b2=0b2x2−2a2x+b2=0⇒b2(x+1x)−2a2=0(Weneedx+1x)LetA=a2+b2+a2−b2&A―=a2+b2−a2−b2x=AA−,x+1x=AA−+A−A=A2+(A―)2AA―=(A+A―)2−2AA―AA―{A+A―=2a2+b2AA―=(a2+b2)2−(a2−b2)2=2b2x+1x=(2a2+b2)2−2(2b2)2b2=2a2b2b2x2−2a2x+b2=0⇒b2(x+1x)−2a2=0⇒b2(2a2b2)−2a2=0⇒0=0(proved)

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