Answer :

Given the roots of the equation are equation (a2 + b2)x2 – 2 (ac + bd)x + (c2 + d2) = 0 are equal.

Comparing with standard quadratic equation ax2 + bx + c = 0


a = (a2 + b2) b = – 2(ac + bd) c = (c2 + d2)


For real and distinct roots: D = 0


Discriminant D = b2 – 4ac = 0


[ – 2(ac + bd)]2 – 4(a2 + b2)(c2 + d2) = 0


4(a2c2 + b2d2 + 2abcd) – 4(a2c2 + a2d2 + b2c2 + b2d2) = 0


using (a + b)2 = a2 + 2ab + b2


4(a2c2 + b2d2 + 2abcd – a2c2 – a2d2 – b2c2 – b2d2) = 0


2abcd – a2d2 – b2c2 = 0


– (2abcd + a2d2 + b2c2) = 0


(ad – bc)2 = 0


ad = bc



Hence proved.


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