Q56 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics
The roots of the equation $x^4+x^3-4 x^2+x+1=0$ are diminished by $h$ so that the transformed equation does not contain $\mathrm{x}^2$ term. If the values of such $\mathrm{h}$ are $\alpha$ and $\beta$, then $12(\alpha-\beta)^2=$
- Correct answerA$35$
- B$25$
- C$105$
- D$115$
Solution
$x^4+x^3-4 x^2+x+1=0$
$$
=(x-1)^2\left(x^2+3 x+1\right)=0
$$
$\therefore$ roots are $1,1, \frac{-3 \pm \sqrt{5}}{2}$
Let $\alpha=1, \beta=1, \gamma=\frac{-3+\sqrt{5}}{2}, \delta=\frac{-3-\sqrt{5}}{2}$
Coefficient of $x^2=\Sigma \alpha \beta$
$$
=\alpha \beta+\beta \gamma+\alpha \gamma+\alpha \delta+\beta \delta+\gamma \delta
$$
when roots are diminished by $h$ coefficient of $x^2=0$
$$
\begin{aligned}
& (\alpha-h)(\beta-h)+(\beta-h)(\gamma-h)+(\alpha-h)(\gamma-h)+ \\
& (\alpha-h)(\delta-h)+(\beta-h)(\delta-h)+(\gamma-h)(\delta-h)=0 \\
& =(1-h)^2+(\gamma-h)(\alpha+\beta-2 h)+(\delta-h)(\alpha+\beta-2 h)+
\end{aligned}
$$
$$
\begin{aligned}
& (\gamma-h)(\delta-h)=0 \\
& =(1-h)^2+(\alpha+\beta-2 h)(\gamma+\delta-2 h)+\left(\gamma \delta+h^2-h(\gamma+\delta)=0\right.
\end{aligned}
$$
Clearly $\alpha+\beta=1+1=2$
$$
\begin{aligned}
& \delta+\gamma=-3 \Rightarrow \gamma \delta=1 \\
& \Rightarrow(1-h)^2+(2-2 h)(-3-2 h)+\left(h^2+1+3 h\right)=0 \\
& \Rightarrow 1+h^2-2 h-6+2 h+4 h^2+h^2+3 h+1=0 \\
& \Rightarrow 6 h^2+3 h-4=0
\end{aligned}
$$
Here if roots now are $\alpha \& \beta$
$$
\begin{aligned}
& \alpha+\beta=\frac{-3}{6}=\frac{-1}{2}, \alpha \beta=\frac{-4}{6}=\frac{-2}{3} \\
& 12(\alpha-\beta)^2=12\left[(\alpha+\beta)^2-4 \alpha \beta\right] \\
& =12\left(\frac{1}{4}+\frac{8}{3}\right)=12\left(\frac{35}{12}\right)=35
\end{aligned}
$$
Hence, the correct answer is option: 'A'.