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

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]

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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