Question

(a-b)x^2+(b-c)x+c-a=0 has equal roots proove that 3a=b+c

Answer 1

oh! that's easy

$(a - b) {x}^{2} + (b - c)x + (c - a) = 0 \\ has \: equal \: roots$

therefore, its discriminant must be equal to zero.

$i.e. \\ {(b - c) }^{2} - 4(a - b)(c - a) = 0 \\ {b}^{2} + {c}^{2} - 2bc - 4(ac - {a}^{2} - bc + ab) = 0 \\ 4 {a}^{2} + {b }^{2} + {c}^{2} - 2bc + 4bc - 4ac - 4ab = 0 \\ {( - 2a) }^{2} + {(b)}^{2} + {(c)}^{2} + 2(b)(c) + 2( - 2a)(c) + 2( - 2a)(b) = 0 \\ {( - 2a + b + c)}^{2} = 0 \\ - 2a + b + c = 0 \\ 2a = b + c$