Find the two values of $x$ for which $3x^2 - 6x - 8 = 0$ - Leaving Cert Mathematics - Question 3 - 2017

To find the values of $x$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, $a = 3$, $b = -6$, and $c = -8$. Now, substituting these values in:

1. Calculate $b^2 - 4ac$:

   $b^2 - 4ac = (-6)^2 - 4(3)(-8) = 36 + 96 = 132$

2. Substitute into the quadratic formula:

   $x = \frac{-(-6) \pm \sqrt{132}}{2(3)} = \frac{6 \pm \sqrt{132}}{6}$

3. Simplify the expression:

   $x = 1 \pm \frac{\sqrt{132}}{6}$

4. Calculate the two values:

   • First value:
     $x_1 = 1 + \frac{\sqrt{132}}{6} \approx 2.9$
   • Second value:
     $x_2 = 1 - \frac{\sqrt{132}}{6} \approx -0.9$

Thus, the two values of $x$ are approximately $2.9$ and $-0.9$.