How do you find the $x$ intercepts for $3 x^{2} - 4 x - 8 = 0$ ?

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Solution

Find the $x$ intercepts for $3 x^{2} - 4 x - 8 = 0$ .
Compare $3 x^{2} - 4 x - 8 = 0$ with the standard quadratic equation $a x^{2} + b x + c = 0$ .
Where, $a = 3, b = - 4, c = - 8$ .
It is known that, the formula to find $x$ -intercepts and where $y = 0$ is $\begin{array}{rcl} x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{array}$ .
Substitute the values from the given equation in the above formula:
$\begin{array}{rcl} x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (3) (- 8)}}{2 (3)} \\ = & \frac{4 \pm \sqrt{16 + 96}}{6} \\ = & \frac{4 \pm \sqrt{112}}{6} \\ = & \frac{4 \pm 4 \sqrt{7}}{6} \\ = & \frac{4 (1 \pm \sqrt{7})}{6} \\ = & \frac{2 (1 \pm \sqrt{7})}{3} \\ = & \frac{2 (1 + \sqrt{7})}{3} and \frac{2 (1 - \sqrt{7})}{3} \end{array}$
Therefore, the $x$ intercepts for $3 x^{2} - 4 x - 8 = 0$ are $\frac{2 (1 + \sqrt{7})}{3} and \frac{2 (1 - \sqrt{7})}{3}$ .

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