Question

$9x^2-4=0$

Answer 1

$$\begin{gathered} \text{Given:} \\ \text{9}{x}^2-4=0 \\ \text{On comparing it with}a{x}^2+bx+c=0,\text{we get:} \\ a=9,b=0\mathrm{and}c=-4 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ \text{=}{\left(0\right)}^2-4\times 9\times (-4) \\ \text{= 144 > 0} \\ \text{Hence, the roots of the equation are real.} \\ \text{Roots}\mathit{\text{α}}\text{and}\mathit{\text{β}}\text{are given by:} \\ \alpha \text{=}\frac{-b+\sqrt{D}}{2a}=\frac{-0+\sqrt{144}}{2\times 9}=\frac{-0+12}{18}=\frac{12}{18}=\frac{2}{3} \\ \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-0-\sqrt{144}}{2\times 9}=\frac{-0-12}{18}=\frac{-12}{18}=\frac{-2}{3} \\ \mathrm{Thus}\text{, the roots of the equation are}\frac{2}{3}\text{and}\frac{-2}{3}\text{.} \end{gathered}$$