Question

Find the quadratic equation whose roots are $\frac{6+\sqrt{3}}{3}\text{and}\frac{6-\sqrt{3}}{3}.$

• A. $x^2-16x+9=0$
• B. $3x^2-12x+11=0$
• C. $6x^2-4x+1=0$
• D. $7x^2-4x+11=0$

Answer 1

The correct option is B $3x^2-12x+11=0$

Let $\alpha \&\beta$ be the roots.
Sum of roots, $\alpha +\beta =\frac{6+\sqrt{3}}{3}+\frac{6-\sqrt{3}}{3}$ $=\frac{12}{3}=4$

Product of roots, $\alpha \beta =\frac{6+\sqrt{3}}{3}\times \frac{6-\sqrt{3}}{3}=\frac{6^2-(\sqrt{3}{)}^2}{9}\alpha \beta =\frac{36-3}{9}=\frac{33}{9}=\frac{11}{3}$

Required equation $x^2-(\alpha +\beta )x+\left(\alpha \beta \right)=0=x^2-4x+\frac{11}{3}=0$

Multiplying by 3

$3x^2-12x+11=0$

$\therefore$ the required quadratic equation is $3x^2-12x+11=0$.