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Asked by nmishra320
| 22nd Jan, 2010,
05:43: PM
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
Topperlearning
Answered by
| 22nd Jan, 2010,
08:48: PM
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