Question

For question 25 please pick 1,2,3 or 4

Answer 1

Answer:

4.

$x=\frac{2}{3} \pm \frac{1}{6}i \sqrt{158}$

Step-by-step explanation:

$18x^2-24x+87=0$

Using Quadratic Formula

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

$x=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:18\cdot \:87}}{2\cdot \:18}$

Solving the discriminant:

$\Delta = \left(-24\right)^2-4\cdot \:18\cdot \:87\\\Delta = 576-6264\\ \Delta=-5688$

$x= \frac{24 \pm \sqrt{-5688} }{36}$

$x= \frac{24 \pm \sqrt{5688i} }{36}$

Once $\sqrt{5688} =\sqrt{2^3\cdot \:3^2\cdot \:79}=6\sqrt{158}$

$x=\frac{24\pm6\sqrt{158}i}{36}$

Dividing the denominator and numerator by 6

$x=\frac{4\pm \sqrt{158}i}{6}$

Now rewrite it:

$x=\frac{4}{6} \pm i\frac{\sqrt{158}}{6}$

$x=\frac{2}{3} \pm i \frac{\sqrt{158}}{6}$

or

$x=\frac{2}{3} \pm \frac{1}{6}i \sqrt{158}$