Question

$3a^2{x}^2+8abx+4b^2=0,a\ne 0$

Answer 1

$$\begin{gathered} \text{Given:} \\ {\text{3a}}^2{x}^2+8abx+4b^2=0 \\ \mathrm{On}\mathrm{c}\text{omparing it with A}{x}^2+Bx+C=0,\text{we get:} \\ A=3a^2,B=8ab\mathrm{and}C=4b^2 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(B^2-4AC) \\ \text{=}{\left(8ab\right)}^2-4\times 3a^2\times 4b^2 \\ {\text{= 16a}}^2{b}^2\text{> 0} \\ \text{Hence, the roots of the equation are real.} \\ \text{Roots}\mathit{\text{α}}\text{and}\mathit{\text{β}}\text{are given by:} \\ \alpha \text{=}\frac{-b+\sqrt{D}}{2a}=\frac{-8ab+\sqrt{16a^2{b}^2}}{2\times 3a^2}=\frac{-8ab+4ab}{6a^2}=\frac{-4ab}{6a^2}=\frac{-2b}{3a} \\ \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-8ab-\sqrt{16a^2{b}^2}}{2\times 3a^2}=\frac{-8ab-4ab}{6a^2}=\frac{-12ab}{6a^2}=\frac{-2b}{a} \\ \mathrm{Thus}\text{, the roots of the equation are}\frac{-2b}{3a}\text{and}\frac{-2b}{a}\text{.} \end{gathered}$$