(x2 + 1)2 – x2 = 0 has ______.

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(x2 + 1)2 – x2 = 0 has no real roots. Explanation: Given equation is (x2 + 1)2 – x2 = 0 ⇒ x4 + 1 + 2x2 – x2 = 0 ......[∵ (a + b)2 = a2 + b2 + 2ab] ⇒ x4 + x2 + 1 = 0 Let x2 = y ∴ (x2)2 + x2 + 1 = 0 y2 + y + 1 = 0 On comparing with ay2 + by + c = 0, we get a = 1, b = 1 and c = 1 Discriminant, D = b2 – 4ac = (1)2 – 4(1)(1) = 1 – 4 = – 3 Since, D < 0 ∴ y2 + y + 1 = 0 i.e., x4 + x2 + 1 = 0 or (x2 + 1)2 – x2 = 0 has no real roots.

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