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The roots of (x – a) (x – a – 1) + (x – a – 1) (x – a – 2) + (x – a) (x – a – 2) = 0, a ∈ R are always :(a) imaginary (b) real and distinct (c) equal (d) rational and equal |
Answer» (b) real and distinct The equation is (x – a) (x – a –1) + (x – a – 1) (x – a – 2) + (x – a) (x – a – 2) = 0. Let (x – a) = y, then the equation becomes y (y – 1) + (y – 1) (y – 2) + y (y – 2) = 0 ⇒ y2 – y + y2 – 3y + 2 + y2 – 2y = 0 ⇒ 3y2 – 6y + 2 = 0 ∴ Discriminant = D = b2 – 4ac = 36 – 4 × 3 × 2 = 36 – 24 = 12 > 0 ∴ Roots are real and distinct. |
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