Question

Form a quadratic polynomial whose zeros are $\frac{3-\sqrt{3}}{5}$ and $\frac{3+\sqrt{3}}{6}$.

Answer 1

The zeros of the polynomial are given as $\frac{3-\sqrt{3}}{5}$ and $\frac{3+\sqrt{3}}{6}$. So,

$\alpha =\frac{3-\sqrt{3}}{5}$ and $\beta =\frac{3+\sqrt{3}}{6}$

The quadratic equation is of the form,

$x^2-\left(\alpha +\beta \right)x+\alpha \beta =0$

$x^2-\left(\frac{3-\sqrt{3}}{5}+\frac{3+\sqrt{3}}{6}\right)x+\left(\frac{3-\sqrt{3}}{5}\right)\left(\frac{3+\sqrt{3}}{6}\right)=0$

$x^2-\left(\frac{6\left(3-\sqrt{3}\right)+5\left(3+\sqrt{3}\right)}{30}\right)x+\frac{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{30}=0$

$30x^2-\left(18-6\sqrt{3}+15+5\sqrt{3}\right)x+\left(9-3\right)=0$

$30x^2-\left(33-\sqrt{3}\right)x+6=0$

Therefore, the required quadratic polynomial is $30x^2-\left(33-\sqrt{3}\right)x+6=0$.