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Prove that both the roots of the equation (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 are real but they are equal only when \(a=b=c.\) |
Answer» For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D > 0, roots are real. (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 ⇒ x2 – (a + b)x + ab + x2 – (b + c)x + bc + x2 – (a + c)x + ac = 0 ⇒ 3x2 - 2(a + b + c)x + ab + bc + ac = 0 ⇒ D = 4(a + b + c)2 – 12(ab + bc + ac) ⇒ D = a2 + b2 + c2 + 2ab + 2ac + 2bc – 3ab – 3bc – 3ac ⇒ D = 1/2 × (2a2 + 2b2 + 2c2 - 2ab – 2ac – 2bc) ⇒ D = 1/2 × ((a – b)2 + (b – c)2 + (c – a)2) Thus, D is always greater than 0, and the roots are real Now, when a = b = c, D = 0, thus the roots are equal when a = b = c. |
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