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Show that the roots of the equation(x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 0 are always real and theycannot be equal unless a = b = c |
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Answer» Root of quadratic equation (x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c) = 0 are equal Means D = b² - 4ac = 0 for this equation, first we should rearrange the equation , (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) ⇒x² - (a + b)x + ab + x² - (b + c)x + bc + x² - (c + a)x + ca⇒3x² - 2(a + b + c)x + (ab + bc + ca) D = {2(a + b + c)}² - 4(ab + bc + ca).3 = 0⇒4{a² + b² + c² + 2(ab + bc + ca)} -12(ab + bc + ca) = 0⇒ a² + b² + c² - ab - bc - ca = 0⇒2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0 ⇒(a - b)² + (b - c)² + (c - a)² = 0 This is possible only when , a = b = c Hence, proved , if roots of given equation are equal then, a = b = c |
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