Q45 / 120 · VITEEE 2006 · Mathematics
The non integer roots of
$x^{4}-3 x^{3}-2 x^{2}+3 x+1=0$
- Correct answerA$\frac{1}{2}(3+\sqrt{13}), \frac{1}{2}(3-\sqrt{13})$
- B$\frac{1}{2}(3-\sqrt{13}), \frac{-1}{2}(3+\sqrt{13})$
- C$\frac{1}{2}(3+\sqrt{17}), \frac{1}{2}(3-\sqrt{17})$
- D$\frac{1}{2}(3-\sqrt{17}), \frac{-1}{2}(3+\sqrt{17})$
Solution
Given $x^{4}-3 x^{3}-2 x^{2}+3 x+1=0$
By using Hit \& trial method, we have $(\mathrm{x}-1)$ is a factor of given equation
$\begin{array}{l}
\therefore(x-1)\left(x^{3}-2 x^{2}-4 x-1\right)=0 \\
\Rightarrow(x-1)\left[x^{3}+x^{2}-3 x^{2}-3 x-x-1\right]=0 \\
\Rightarrow(x-1)\left[x^{2}(x+1)-3 x(x+1)-1(x+1)\right]=0 \\
\Rightarrow(x-1)(x+1)\left(x^{2}-3 x-1\right)=0 \\
\therefore x=1,-1 \text { or } x^{2}-3 x-1=0 \\
\text { Now } x^{2}-3 x-1=0 \\
\Rightarrow x=\frac{3 \pm \sqrt{9+4}}{2}
\end{array}$
$\left[\because \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^{2}-4 \mathrm{ac}}}{2 \mathrm{a}}\right]$
$\Rightarrow \mathrm{x}=\frac{3 \pm \sqrt{13}}{2}$
$\therefore$ non-integer roots of given equation are
$\frac{1}{2}(3+\sqrt{13}), \frac{1}{2}(3-\sqrt{13})$
Hence, the correct answer is option: 'A'.