Question

For the equation $6x^4-13x^3-35x^2-x+3=0,2+\sqrt{3}$ is one root of it, then the quadratic equation, whose roots are the rational roots of the above equation, is:

• A. $6x^2+11x+3=0$
• B. $6x^2-11x+3=0$
• C. $6x^2+11x-3=0$
• D. $6x^2-11x-3=0$

Answer 1

The correct option is A $6x^2+11x+3=0$

If one root of $6x^4-13x^3-35x^2-x+3=0$ is $2+\sqrt{3}$, then the other root will be $2-\sqrt{3}$
(As irrational roots occurs in conjugate pair)
Let rational roots be $\alpha ,\beta$
Then sum of roots $=\alpha +\beta +2+\sqrt{3}+2-\sqrt{3}=\frac{13}{6}$
$\Rightarrow \alpha +\beta =-\frac{11}{6}$ ....(1)
Product of roots $=\alpha \beta (2+\sqrt{3})(2-\sqrt{3})=\frac{3}{6}$
$\Rightarrow \alpha \beta =\frac{3}{6}$ ....(2)
From (1) and (2), we get the required equation
$x^2-(-\frac{11}{6})x+\frac{3}{6}=0\Rightarrow 6x^2+11x+3=0$
Hence, option A is correct.