Q. 124.4( 7 Votes )

Prove that the equation x2(a2 + b2) + 2x(ac + bd) + (c2 + d2)= 0 has no real root, if ad ≠ bc.

Answer :

x2(a2 + b2) + 2x(ac + bd) + (c2 + d2)= 0


d = b2 – 4ac


d = (2ac + 2bd)2 – 4 (a2 + b2) (c2 + d2)


d = 4a2c2 + 4b2d2 + 8abcd – 4 [a2 (c2 + d2) + b2 (c2+d2)]


d = 4a2c2 + 4b2d2 + 8abcd – 4 [a2c2 + a2d2 + b2c2 + b2d2]


d = 4a2c2 + 4b2d2 + 8abcd – 4a2c2 – 4a2d2 – 4b2c2 – 4b2 d2


d = 8abcd – 4a2d2 – 4b2c2


d = 8abcd – 4(a2d2 + b2 c2)


d = –4 (a2 d2 + b2c2 – 2abcd)


d = –4 [(ad + bc)2]


For ad ≠ bc


d= –4 × [value of (ad + bc)2]


d is always negative


So, d < 0


The given equation has no real roots.


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