Question

The roots of the equation $4\sqrt{3}{x}^2+5x-2\sqrt{3}=0$ are
(a) $\frac{2\sqrt{3}}{3},\frac{-\sqrt{3}}{4}$
(b) $\frac{-2\sqrt{3}}{3},\frac{\sqrt{3}}{4}$
(c) $\frac{\sqrt{3}}{3},\frac{-\sqrt{3}}{4}$
(d) $\frac{-\sqrt{3}}{3},\frac{\sqrt{3}}{4}$

Answer 1

(b) $\frac{-2\sqrt{3}}{3},\frac{\sqrt{3}}{4}$

$$\begin{gathered} \text{Given:} \\ \text{4}\sqrt{3}{x}^2+5x-2\sqrt{3}=0 \\ \Rightarrow \text{4}\sqrt{3}{x}^2+8x-3x-2\sqrt{3}=0 \\ \text{⇒ 4}x(\sqrt{3}x+2)-\sqrt{3}(\sqrt{3}x+2)=0 \\ \text{⇒}(\sqrt{3}x+2)(4x-\sqrt{3})=0 \\ \text{⇒}x=\frac{\sqrt{3}}{4}\text{or}x=\frac{-2}{\sqrt{3}}=\frac{-2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-2\sqrt{3}}{3} \\ \text{Thus, the roots of the equation are}\frac{\sqrt{3}}{4}\text{and}\frac{-2\sqrt{3}}{3}\text{.} \end{gathered}$$