Question

pls help me out in this question
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Answer 1

Given that roots of the equation
$a {(b - c)}x^{2} + b(c - a)x + c(a - b) = 0$
are equal.


For a quadratic equation with equal roots the Discriminant equals 0.


Therefore,

${(b(c - a))}^{2} - 4(a(b - c))(c(a - b)) = 0 \\ {b}^{2} ( {c}^{2} - 2ac + {a ^{2} ) } - 4(ab - ac)(ac - bc) = 0 \\ {b}^{2} {c}^{2} - 2a {b}^{2} c + a ^{2} {b}^{2} - 4( {a}^{2} bc - a {b}^{2} c - {a}^{2} {c}^{2} + ab {c}^{2} ) = 0 \\ {b}^{2} {c}^{2} - 2a {b}^{2} c + a ^{2} {b}^{2} - 4a ^{2} {b} c + 4a {b}^{2} c + 4 {a}^{2} {c}^{2} - 4ab {c}^{2} = 0 \\ {b}^{2} {c}^{2} + 2a {b}^{2} c + {a}^{2} {b}^{2} - 4 {a}^{2} bc + 4 {a}^{2} {c}^{2} - 4ab {c}^{2} = 0 \\ (bc) ^{2} + 2(ab)(bc) + (ab)^{2} + 2 ( - 2ac)(ab) + ( - 2ac) ^{2} + 2( - 2ac)(bc) = 0 \\ (bc + ab - 2ac) ^{2} = 0 \\ bc + ab - 2ac = 0 \\ bc + ab = 2ac \\ \frac{bc}{abc} + \frac{ab}{abc} = \frac{2ac}{abc} \\ \\ \frac{1}{a} + \frac{1}{c} = \frac{2}{b}$



Done. (Sorry for the delay, Latex takes time.)