Question

Solve the following quadratic equation using quadratic formula: $3a^2{x}^2+abx-2b^2=0$

• A. $\{\frac{2b}{3a},-\frac{b}{a}\}$
• B. $\{\frac{2b}{3a},\frac{b}{a}\}$
• C. $\{-\frac{2b}{3a},\frac{-b}{a}\}$
• D. None of these

Answer 1

The correct option is A $\{\frac{2b}{3a},-\frac{b}{a}\}$

$3a^2{x}^2+abx-2b^2=0$

$\Rightarrow$ Here, $A=3a^2,B=ab,C=-2b^2$

$\Rightarrow$ $x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$

$=\frac{-ab\pm \sqrt{(ab{)}^2-4(3a^2\left)\right(-2b^2)}}{2\left(3a^2\right)}$

$=\frac{-ab\pm \sqrt{a^2{b}^2+24a^2{b}^2}}{6a^2}$

$=\frac{-ab\pm \sqrt{25a^2{b}^2}}{6a^2}$

$=\frac{-ab\pm 5ab}{6a^2}$

$\Rightarrow$ $x=\frac{-ab+5ab}{6a^2}$ and $x=\frac{-ab-5ab}{6a^2}$

$\therefore$ $x=\frac{2b}{3a}$ and $x=-\frac{b}{a}$