Question

Question 1 (ii)
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(ii) $3x^2-4\sqrt{3}x+4=0$

Answer 1

(ii) $3x^2-4\sqrt{3}x+4=0$
Comparing it with $a{x}^2+bx+c=0$, we get
$a=3,b=-4\sqrt{3}andc=4$
$Discriminant=b^2-4ac$
$=(-4\sqrt{3}{)}^2-4(3\left)\right(4)$
$=48-48=0Since{b}^2-4ac=0,$
Real roots exist for the given equation and they are equal to each other. The roots will be $\frac{-b}{2a}and\frac{-b}{2a}$.
$\Rightarrow$ $\frac{-b}{2a}$= - $\frac{(-4\sqrt{3})}{2\times 3}$ = $\frac{4\sqrt{3}}{6}$ = $\frac{2\sqrt{3}}{3}$ = $\frac{2}{\sqrt{3}}$
Therefore, the roots are $\frac{2}{\sqrt{3}}and\frac{2}{\sqrt{3}}$.