Question

If one root of the equation $x^4-9x^3+27x^2-29x+6=0$ is $2-\sqrt{3}$, then the number of rational root(s) of the equation are

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

Let $P\left(x\right)=x^4-9x^3+27x^2-29x+6$.
Since the coefficients are rational and $2-\sqrt{3}$ is one of the root,
hence $2+\sqrt{3}$ is also a root of $P\left(x\right)=0$

Let the remaining two roots be $\alpha ,\beta$.
Sum of roots $=\alpha +\beta +2+\sqrt{3}+2-\sqrt{3}=9$
$\Rightarrow \alpha +\beta =5\cdots \left(1\right)$

Products of roots $=\alpha \beta (2+\sqrt{3})(2-\sqrt{3})=6$
$\Rightarrow \alpha \beta =6\cdots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get
$\alpha =2,\beta =3$ or $\alpha =3,\beta =2$
Hence, the roots of the given equation are $2,3,2\pm \sqrt{3}$