• # question_answer If one root of the equation $a{{x}^{2}}+bx+c=0$be $n$ times the other root, then A) $n{{a}^{2}}=bc{{(n+1)}^{2}}$ B) $n{{b}^{2}}=ac{{(n+1)}^{2}}$ C) $n{{c}^{2}}=ab{{(n+1)}^{2}}$ D) None of these

Solution :

Let the roots are $\alpha$and $n\alpha$ Sum of roots$\alpha +n\alpha =-\frac{b}{a}$Þ$\alpha =-\frac{b}{a(n+1)}$ .....(i) and product, $\alpha .n\alpha =\frac{c}{a}$Þ ${{\alpha }^{2}}=\frac{c}{na}$ ....(ii) From (i) and (ii), we get Þ ${{\left[ -\frac{b}{a(n+1)} \right]}^{2}}=\frac{c}{na}$Þ $\frac{{{b}^{2}}}{{{a}^{2}}{{(n+1)}^{2}}}=\frac{c}{na}$ $\Rightarrow$ $n{{b}^{2}}=ac{{(n+1)}^{2}}$. Note: Students should remember this question as a fact.

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