If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal, then prove that 2b = a + c.

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Anupam Gupta

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General form is Ax^2 + Bx + C= 0 if A + B + C = 0 then zeroes of polynomial are 1 and C/A here A + B + C = b -c +c - a + a - b = 0 therefore zeroes are 1 and (a - b) / (b - c) its given that they are equal â?´ (a - b) / (b - c) = 1 â??a - b = b - c 2b = a + c hence proved
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Math Magician

For equal roots the discriminant of the quadratic equation should be zero. Hence, (c-a) ^2-4(b-c)(a-b)=0 After simplification we get, a^2+c^2-4b^2+2ac-4bc-4ab=0 Hence, (a-2b+c)^2=0 implies, a+c=2b
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