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If (a2 + b2) x2 + 2 (ac + bd) x + c2 + d2 = 0 has no real roots, then A. ad = bc B. ab = cd C. ac = bd D. ad ≠ bc |
Answer» Since the equation (a2 + b2) x2 + 2 (ac + bd) x + c2 + d2 = 0 has no real root D < 0 b2 – 4ac < 0 b2 < 4ac Here a = (a2 + b2), b = 2 (ab + bd), c = c2 + d2 4(ac + bd)2 – 4 (a2 + b2)(c2 + d2) < 0 4a2c2 + 4b2d2 + 8abcd – 4(a2c2 + b2c2 + a2d2 + b2d2) < 0 – 4(a2d2 + b2c2 – 2abcd) < 0 – 4(ad + bc)2< 0 ∴ d is always negative And ad \(\neq\) bc |
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