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Question

If the roots of the equation (a2+b2)x22(ac+bd)x+(c2+d2)=0 are equal, then prove that ab=cd.

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Solution

(a2+b2)x22(ac+bd)x+(c2+d2)=0
Comparing with Ax2+Bx+C=0
A=(a2+b2), B=2(ac+bd), C=(c2+d2)
For equal roots, B24AC=0
(2(ac+bd))24(a2+b2)(c2+d2)=0
4(a2c2+2abcd+b2d2)4(a2c2+a2d2+b2c2+b2d2)=0, divide by 4 to get
a2c2+2abcd+b2d2a2c2a2d2b2c2b2d2=0
2abcda2d2b2c2=0, Multiply by -1 to get,
a2d22abcd+b2c2=0
(adbc)2=0, Take square root on both sides to get,
adbc=0
ad=bc
ab=cd


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