InterviewSolution
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InterviewSolution
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In the quadratic equation `ax^2 + bx + c = 0`, if `Delta = b^2-4ac and alpha + beta, alpha^2 + beta^2, alpha^3 + beta^3` are in GP. where `alpha, beta` are the roots of `ax^2 + bx + c =0`, then |
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Answer» `ax^2 + bx+c = 0` `alpha and beta` are roots `alpha + beta = -b/a` `alpha*beta= c/a` given, `/_ = b^2- 4ac` & `alpha+ beta , alpha^2+beta^2, alpha^3+ beta^3` are in GP. :. the condition will be `(alpha^2 + beta^2)^2 = (alpha+ beta)*(alpha^3+ beta^3)` solving further: `[(alpha+beta)^2 - 2alpha beta]^2 = (alpha+beta)*[(alpha+beta)^3 - 3alpha beta (alpha+ beta)]` `[b^2/a^2 - 2c/a]^2 = (-b/a)[-b^3/a^3 +3*c/a*b/a]` `[(b^2-2ac)/a^2]^2 = (-b/a)[(-b^3+3abc)/a^3]` `[(b^2-2ac)/a^2] = b^2/a^4[b^2 - 3ac]` `(b^2 - 2ac)^2/cancel(a^4) = b^2/cancel(a^4)[b^2 -3ac]` `cancel(b^4) + 4a^2c^2 - 4b^2ac= cancel(b^4) - 3b^2ac` `4a^2c^2 - b^2ac = 0` `-ac(b^2 - 4ac) = 0` `-ac. /_ = 0` so` /_ = 0` `c/_ = 0` option (c) is correct |
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