Question

If the roots of the equation (a-b)x^2+(b-c)x+( c-a)=0 are equal then prove that 2a= b+c

Answer 1

$\mathfrak{\huge{Answer:-}} \\ \bold{(a - b){x}^{2} + (b - c)x + (c - a) = 0} \\ a = (a - b) \: b = (b - c) \: c = (c - a)\\ It \: has \: equal \: roots, \: then \\Discriminant = 0 \\ {b}^{2} - 4ac = 0 \\ {(b - c)}^{2} - 4(a - b)(c - a) = 0 \\ {b}^{2} + {c}^{2} - 2bc - 4[a(c - a) - b(c - a)] = 0 \\ {b}^{2} + {c}^{2} - 2bc - 4(ac - {a}^{2} - bc + ab) = 0 \\ {b}^{2} + {c}^{2} - 2bc - 4ac + 4 {a}^{2} + 4bc - 4ab = 0 \\ 4 {a}^{2} + {b}^{2} + {c}^{2} - 4ab - 2bc + 4bc - 4ac = 0 \\ 4 {a}^{2} + {b}^{2} + {c}^{2} - 4ab + 2bc - 4ab = 0 \\ {(2a)}^{2} + {(b)}^{2} + {(c)}^{2} - 2(2a)(b) + 2(b)(c) - 2(c)(2a)\\ Using \: the \: identity :- \: \: {a}^{2} + {b}^{2} + {c}^{2} - 2ab + 2bc - 2ca = {(a - b - c)}^{2} \\ a = 2a, \: b = b ,\: c = c \\ {(2a - b - c)}^{2} = 0 \\ 2a - b - c = \sqrt{0} \\ 2a - b - c = 0 \\ 2a = b + c \\ Hence \: proved$