Question

Find the zeroes of the quadratic polynomial $\sqrt{3x^2}-8x-4\sqrt{3}$.

Answer 1

Factorise $\sqrt{3x^2}-8x-4\sqrt{3}=0$ to find zeroes of quadratic polynomial:
Let the polynomial $p\left(x\right)=\sqrt{3x^2}-8x-4\sqrt{3}$.

Definition:

Any real number $\text{'}a\text{'}$ is root or zero of a polynomial $p\left(x\right)$ if and only if $p\left(a\right)=0$.

On setting $p\left(a\right)=0$.

$\begin{array}{cc}\Rightarrow & \sqrt{3a^2}-8a+4\sqrt{3}=0\\ \Rightarrow & \sqrt{3a^2}-2a-6a+4\sqrt{3}=0\\ \Rightarrow & a\left(\sqrt{3}a-2\right)-2\sqrt{3}\left(\sqrt{3}a-2\right)=0\\ \Rightarrow & \left(\sqrt{3}a-2\right)\left(a-2\sqrt{3}\right)=0\end{array}$

On further simplification,

$\begin{array}{cccc}\Rightarrow & \left(\sqrt{3}a-2\right)=0& \text{and}& \left(a-2\sqrt{3}\right)=0\\ \Rightarrow & a=\frac{2}{\sqrt{3}}& \text{and}& a=2\sqrt{3}\end{array}$

Thus, the zeroes of polynomial $\sqrt{3x^2}-8x-4\sqrt{3}$ are $\frac{2}{\sqrt{3}},2\sqrt{3}$.