Chapter 7.3, Problem 10WE

To determine

Find the nature of roots of the equation $\sqrt{3}{x}^2-4\sqrt{3}=0$ without solving it.

Answer to Problem 10WE

The equation has two real and unequal roots.

Explanation of Solution

Given information: The given quadratic equation is,

  $\sqrt{3}{x}^2-4\sqrt{3}=\mathrm{0....}\left(\text{Given}\right)$

Calculation: The given equation is,

  $\sqrt{3}{x}^2-4\sqrt{3}=\mathrm{0....}\left(\text{1}\right)$

Now if $\left(D\right)$ is the discriminant of quadratic equation (1) then using the formula $D=b^2-4ac$ ,

  $\begin{array}{c}D={\left(0\right)}^2-4\left(\sqrt{3}\right)\left(-4\sqrt{3}\right)\mathrm{....}\left\{\begin{array}{l}a=\sqrt{3},b=0\\ \text{and }c=-4\sqrt{3}\end{array}\right\}\\ =0+48\\ =48\\ >0\end{array}$

Since the value of discriminant $\left(D\right)$ is $48$ which is greater than zero, therefore the roots are real and unequal to each other.