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Question
If ad ≠ bc, then prove that the equation (a2 + b2) x2 + 2 (ac + bd) x + (c2 + d2) = 0 has no real roots.
Solution
The given equation is (a2 + b2)x2 + 2 (ac + bd)x + (c2 + d2) = 0
We know, D = b2 − 4ac
Thus,
D=[2(ac + bd)2] −4(a2 + b2)(c2 + d2)
=[4(a2c2 + b2d2 + 2abcd)] −4(a2 + b2)(c2 + d2)
=4[(a2c2 + b2d2 + 2abcd) − (a2c2 + a2d2 + b2c2 + b2d2)]
=4[a2c2 + b2d2 + 2abcd − a2c2 − a2d2 − b2c2 − b2d2]
=4[2abcd − b2c2 − a2d2]
=−4[a2d2 + b2c2 − 2abcd]
=−4[ad − bc]2
But we know that ad ≠ bc
Therefore,
(ad−bc) ≠ 0
⇒(ad − bc)2 > 0
⇒−4(ad − bc)2 <0
⇒D < 0
Hence, the given equation has no real roots.
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