# How do you solve |x^2 - 3x| = 18?

Apr 4, 2016

Only real roots are $x = 6$ and $x = - 3$.

Complex roots are $x = \frac{3}{2} + i \frac{\sqrt{63}}{2}$ and $x = \frac{3}{2} - i \frac{\sqrt{63}}{2}$

#### Explanation:

As $| {x}^{2} - 3 x | = 18$, we have either ${x}^{2} - 3 x = 18$ i.e. ${x}^{2} - 3 x - 18 = 0$ or

${x}^{2} - 3 x = - 18$ i.e. ${x}^{2} - 3 x + 18 = 0$

For solving ${x}^{2} - 3 x - 18 = 0$ by factorizing, we split middle term as

${x}^{2} + 3 x - 6 x - 18 = 0$ or $x \left(x + 3\right) - 6 \left(x + 3\right) = 0$ or $\left(x - 6\right) \left(x + 3\right) = 0$ i.e. $x = 6$ or $x = - 3$.

As determinant given by ${b}^{2} - 4 a c$ for ${x}^{2} - 3 x + 18 = 0$

is ${\left(- 3\right)}^{2} - 4 \cdot 1 \cdot 18 = 9 - 72 = - 63$,

Hence, we do not have any real roots of ${x}^{2} - 3 x + 18 = 0$

Complex roots for this are given by $x = \frac{3 \pm \sqrt{- 63}}{2}$ or

$x = \frac{3}{2} + i \frac{\sqrt{63}}{2}$ or $x = \frac{3}{2} - i \frac{\sqrt{63}}{2}$