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Asked by nmishra320 | 22nd Jan, 2010, 05:43: PM

Expert Answer:

Dear Student

Solution of ax^2+ bx+c=0  ==> (b +/- sqrt (b^2-4ac))/2a

Basec on the same formula

4x2-2(a2+b2)x+a2b2.=0

X={   -(2(a2+b2)) +/-  [ 4(a2+b2)2 - 4X 4Xa2b2 ] } / (2X4)

={   -(2(a2+b2)) +/-2 [ a4+b4-2a2b2 ] } / 8

={   -(2(a2+b2)) +/-2 [ (a2-b2) ] } / 8

={   -(a2+b2) +/- [ (a2-b2)2 ] } / 4

={   -(a2+b2) +/-  (a2-b2) } / 4

==> the two solutions are 

X=   {   -a2-b2 + a2-b2} / 4    ;    {   -a2-b2 -  a2+b2 } / 4

X= -b2 / 2   ;   -a2 / 2   ;  

Regards

Team

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Answered by  | 22nd Jan, 2010, 08:48: PM

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