Solve the equation 3x^{2} - 4x + frac{20}{3} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given quadratic equation is $3x^{2} - 4x + \frac{20}{3} = 0$.Multiplying the entire equation by $3$,we get $9x^{2} - 12x + 20 = 0$.Comparing t

Vedclass · Accepted Answer

$\frac{2}{3} \pm \frac{4}{3}i$

Vedclass · Answer

(A) The given quadratic equation is $3x^{2} - 4x + \frac{20}{3} = 0$.Multiplying the entire equation by $3$,we get $9x^{2} - 12x + 20 = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we have $a = 9$,$b = -12$,and $c = 20$.The discriminant $D$ is given by $D = b^{2} - 4ac = (-12)^{2} - 4(9)(20) = 144 - 720 = -576$.Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$,we get:$x = \frac{-(-12) \pm \sqrt{-576}}{2(9)} = \frac{12 \pm \sqrt{576}i}{18} = \frac{12 \pm 24i}{18}$.Simplifying the expression,we get $x = \frac{12}{18} \pm \frac{24}{18}i = \frac{2}{3} \pm \frac{4}{3}i$.

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

Similar Questions

If $AM$ and $GM$ of roots of a quadratic equation are $5$ and $4$,respectively,then the quadratic equation is

The equation $16x^4 + 16x^3 - 4x - 1 = 0$ has a multiple root. If $\alpha, \beta, \gamma, \delta$ are the roots of this equation,then $\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} =$

Solve the equation $\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0$.

Let $a$ be an integer selected at random from the set $\{0, 1, 2, 3, ..., 9\}$. The probability that the equation $ax^2 - ax + 1 = 0$ has real roots is ...

The product of the roots of the equation $9x^{2}-18|x|+5=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module