If -1 is a twice repeated root of the equation $a x^3+b x^2+c x+1=0$, then

If -1 is a twice repeated root of the equation $a x^3+b x^2+c x+1=0$, then
  1. $\mathrm{b}=2 \mathrm{a}+1, \mathrm{c}=\mathrm{a}+1$
  2. $\mathrm{b}=2 \mathrm{a}+1, \mathrm{c}=\mathrm{a}-2$
  3. $\mathrm{b}=2 \mathrm{a}+1, \mathrm{c}=\mathrm{a}+2$
  4. $b=2 a-1, c=a+2$

Solution

Let roots of equation be $-1,-1, \alpha$ Now, $\alpha=\frac{-1}{a}$ and $-1-1-\frac{1}{a}=\frac{-b}{a}$ $\Rightarrow b=2 a+1$ and $-a+b-c+1=0 \Rightarrow c=-a+b+1$ $\Rightarrow c=a+2$

Asked in: AP EAMCET 2023 (15 May Shift 2)

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