Question

$3x^2-4x+\frac{20}{3}=0$

Answer 1

Given: $3x^2-4x+\frac{20}{3}=0$
Comparing the given equation with the general form of the quadratic equation $a{x}^2+bx=c=0$, we get $a=3,b=-4$ and $c=\frac{20}{3}$.


Substituting these values in $\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}$, we get:

$\Rightarrow \alpha =\frac{4+\sqrt{16-4\times 3\times {\displaystyle \frac{20}{3}}}}{6}$ and $\beta =\frac{4-\sqrt{16-4\times 3\times {\displaystyle \frac{20}{3}}}}{6}$

$\Rightarrow \alpha =\frac{4+\sqrt{-64}}{6}$ and $\beta =\frac{4-\sqrt{-64}}{6}$

$\Rightarrow \alpha =\frac{4+8i}{6}$ and $\beta =\frac{4-8i}{6}$

$\Rightarrow \alpha =\frac{2+4i}{3}$ and $\beta =\frac{2-4i}{3}$


Hence, the roots of the equation are $\frac{2\pm 4i}{3}$.