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Question
If the roots of the equation (a2 + b2)x2 − 2 (ac + bd)x + (c2 + d2) = 0 are equal, prove that `a/b=c/d`.
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Solution
The given quadric equation is (a2 + b2)x2 − 2 (ac + bd)x + (c2 + d2) = 0, and roots are real
Then prove that `a/b=c/d`.
Here,
a = (a2 + b2), b = -2 (ac + bd) and c = (c2 + d2)
As we know that D = b2 - 4ac
Putting the value of a = (a2 + b2), b = -2 (ac + bd) and c = (c2 + d2)
D = b2 - 4ac
= {-2(ac + bd)}2 - 4 x (a2 + b2) x (c2 + d2)
= 4(a2c2 + 2abcd + b2 + d2) - 4(a2c2 + a2d2 + b2c2 + b2d2)
= 4a2c2 + 8abcd + 4b2d2 - 4a2c2 - 4a2d2 - 4b2c2 - 4b2d2
= -4a2d2 - 4b2c2 + 8abcd
= -4(a2d2 + b2c2 - 2abcd)
The given equation will have real roots, if D = 0
-4(a2d2 + b2c2 - 2abcd) = 0
a2d2 + b2c2 - 2abcd = 0
(ad)2 + (bc)2 - 2(ad)(bc) = 0
(ad - bc)2 = 0
Square root both sides we get,
ad - bc = 0
ad = bc
`a/b=c/d`
Hence `a/b=c/d`
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