Question

Find a quadratic polynomial whose zeros are $-\sqrt{3}$ and $\frac{-7}{\sqrt{3}}$.

Answer 1

It is given that $-\sqrt{3}$ and $\frac{-7}{\sqrt{3}}$ are the zeros of the polynomial, so,

$\alpha =-\sqrt{3}$ and $\beta =\frac{-7}{\sqrt{3}}$

The quadratic equation is of the form,

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

$x^2-\left(-\sqrt{3}-\frac{7}{\sqrt{3}}\right)x+\left(-\sqrt{3}\right)\left(-\frac{7}{\sqrt{3}}\right)=0$

$x^2-\left(\frac{-3-7}{\sqrt{3}}\right)x+7=0$

$x^2+\frac{10}{\sqrt{3}}x+7=0$

$\sqrt{3}{x}^2+10x+7\sqrt{3}=0$

Therefore, the required quadratic polynomial is $\sqrt{3}{x}^2+10x+7\sqrt{3}=0$.