Question

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.

Equation: $3x^2-4\sqrt{3}x+4=0$

Answer 1

Step 1: Evaluate the discriminant to identify if the equation has real roots

We know that discriminant $D=b^2-4ac$

The condition to have real roots in a quadratic equation is $D\ge 0$.

Given equation, $3x^2-4\sqrt{3}x+4=0$, where $a=3,b=-4\sqrt{3},c=4$

So, the discriminant will be,

$$\begin{gathered} \Rightarrow D={(-4\sqrt{3})}^2-(4\times 3\times 4) \\ \Rightarrow D=48-48 \\ \Rightarrow D=0 \end{gathered}$$

As $D=0$, so the given equation satisfies the required condition to have real roots.

So, $3x^2-4\sqrt{3}x+4=0$ does have real roots.

Step 2: Evaluate the roots of the equation

Solve by middle-term splitting,

$$\begin{gathered} \Rightarrow 3x^2-4\sqrt{3}x+4=0 \\ \Rightarrow 3x^2-2\sqrt{3}x-2\sqrt{3}x+4=0 \\ \Rightarrow \sqrt{3}x(\sqrt{3}x-2)-2(\sqrt{3}x-2)=0 \\ \Rightarrow (\sqrt{3}x-2)(\sqrt{3}x-2)=0 \\ \Rightarrow x=\frac{2}{\sqrt{3}} \end{gathered}$$

Hence, $3x^2-4\sqrt{3}x+4=0$ has real roots which are $\frac{2}{\sqrt{3}}$ and $\frac{2}{\sqrt{3}}$.