If -1 is a twice repeated root of the equation $a\left(x^3+x^2\right)+$ $\mathrm{bx}+\mathrm{c}=0$, then $\mathrm{a}: \mathrm{b}: \mathrm{c}=$

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Aicharya · Accepted Answer

Since $x=1$ is a twice repeated root of $f(x)=a x^3+a x^2$ $+b x+c=0$. Hence $f(-1)=0$ and $f^{\prime}(-1)=0$ Now $f(-1)=0$ $ \Rightarrow-a+a+b+c=0 \Rightarrow c=b $ Now $f^{\prime}(x)=3 a x^2+2 a x+b$ $ \begin{aligned} & \Rightarrow f^{\prime}(-1)=3 a-2 a+b \ & \Rightarrow a+b=0 \ & \Rightarrow a=-b \ & 	herefore a: b: c=(-b):(b):(b) \ & =-1: 1: 1 \end{aligned} $

If -1 is a twice repeated root of the equation $a\left(x^3+x^2\right)+$ $\mathrm{bx}+\mathrm{c}=0$, then…

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