Q54 / 160 · TS EAMCET 2020 (09 Sep Shift 2) · Mathematics

If the roots of $x^4+x^3-4 x^2+x+1=0$ are diminished by $\alpha$ or $\beta$, then the equation with the diminished roots does not contain $x^2$ term. Then $12(\alpha-\beta)^2=$

  • Correct answer
    A
    $35$
  • B
    $105$
  • C
    $96$
  • D
    $115$

Solution

Let the roots of equation $x^4+x^3-4 x^2+x+1=0$ are diminished by $h$, so the equation becomes $(x+h)^4+(x+h)^3-4(x+h)^2+(x+h)+1=0$ Now the coefficient of $x^2$ is $6 h^2+3 h-4$ must be zero $\therefore \quad 6 h^2+3 h-4=0$ having roots $\alpha$ and $\beta$ So, $(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta$ $\begin{aligned} & =\left(\frac{-3}{6}\right)^2-4\left(-\frac{4}{6}\right) \\ & =\frac{1}{4}+\frac{8}{3}=\frac{3+32}{12}=\frac{35}{12}\end{aligned}$ Therefore $12(\alpha-\beta)^2=35$

Hence, the correct answer is option: 'A'.

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