Question

# Solve the equation : $3{x^2} - 4x + \frac{{20}}{3} = 0$

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Hint:- Use Quadratic formula ($x = \dfrac{{ - b \pm \sqrt D }}{{2a}} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$) to solve this given equation.

Here the given equation is $3{x^2} - 4x + \dfrac{{20}}{3} = 0$
This given equation can also be written as $9{x^2} - 12x + 20 = 0$
On comparing this equation with $a{x^2} + bx + c = 0$
We obtain $a = 9,b = - 12$ and $c = 20$
Therefore, the discriminant of the given equation is
$D = {b^2} - 4ac = {( - 12)^2} - 4 \times 9 \times 20$
$= 144 - 720 = - 576$
Therefore, the required solutions are
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

$= \dfrac{{ -(-12) \pm \sqrt { - 576} }}{{2 \times 9}} = \dfrac{{ -(-12) \pm \sqrt {576} i}}{{18}} \\ = \dfrac{{12 \pm 24i}}{{18}} = \dfrac{{6(2 \pm 4i)}}{{18}} \\ = \dfrac{{2 \pm 4i}}{3} \\$
$\therefore x = \dfrac{2}{3} \pm \dfrac{4}{3}i$ Ans.

Note:- All the quadratic equation cannot be factorize , so we have to apply the formula $x = \dfrac{{ - b \pm \sqrt D }}{{2a}} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to solve such type of questions.