Question

If $2+\sqrt{3}$ is a root of $6x^4-13x^3-35x^2-x+3=0$, then

• A. only two roots are rational
• B. only two roots are real but not equal
• C. only three roots are rational
• D. no root is rational

Answer 1

The correct option is A only two roots are rational

The equation $6x^4-13x^3-35x^2-x+3=0$ has all rational coefficients.
As $2+\sqrt{3}$ is a root, its conjugate $2-\sqrt{3}$ is also a root of the equation in order to have rational coefficients.
Hence, $(x-(2+\sqrt{3}\left)\right)(x-(2-\sqrt{3}\left)\right)$ is a factor of the equation.
$\Rightarrow (x^2-4x+1)$ is a factor of $6x^4-13x^3-35x^2-x+3=0$
$\Rightarrow 6x^4-13x^3-35x^2-x+3=(x^2-4x+1)(a{x}^2+bx+c)$
On comparing the coefficients, we get
$a=6,b=11,c=3$
Hence, $6x^4-13x^3-35x^2-x+3=(x^2-4x+1)(6x^2+11x+3)$
$\Rightarrow 6x^4-13x^3-35x^2-x+3=(x^2-4x+1)(2x+3)(3x+1)$
Hence the roots are $2+\sqrt{3},2-\sqrt{3},\frac{-1}{3},\frac{-3}{2}$
So, two roots are rational.