If one root of the equation ax^2 + bx + c = 0 | vedclass

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(B) Let the roots be $\alpha$ and $n\alpha$. Sum of roots: $\alpha + n\alpha = -\frac{b}{a} \implies \alpha(n + 1) = -\frac{b}{a} \implies \alpha = -\

Vedclass · Accepted Answer

$nb^2 = ac(n + 1)^2$

Vedclass · Answer

(B) Let the roots be $\alpha$ and $n\alpha$. Sum of roots: $\alpha + n\alpha = -\frac{b}{a} \implies \alpha(n + 1) = -\frac{b}{a} \implies \alpha = -\frac{b}{a(n + 1)}$ ... $(i)$ Product of roots: $\alpha \cdot n\alpha = \frac{c}{a} \implies n\alpha^2 = \frac{c}{a} \implies \alpha^2 = \frac{c}{na}$ ... $(ii)$ Substituting $(i)$ into $(ii)$: $\left( -\frac{b}{a(n + 1)} \right)^2 = \frac{c}{na}$ $\frac{b^2}{a^2(n + 1)^2} = \frac{c}{na}$ $nb^2 = ac(n + 1)^2$.

If one root of the equation $ax^2 + bx + c = 0$ is $n$ times the other root,then:

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