How do you solve using completing the square method x^2 - 3x = 18?

Apr 6, 2016

$x = - 3$
$x = 6$

Explanation:

Note: The goal of completing the square is to create a perfect trinomial in the form of

$\left({a}^{2} + 2 a b + {b}^{2}\right) = {\left(a + b\right)}^{2}$ $\text{ " } \mathmr{and}$
$\left({a}^{2} - 2 a b + {b}^{2}\right) = {\left(a - b\right)}^{2}$

Given:

${x}^{2} - 3 x = 18$

Step 1: The leading coefficient is 1, we can proceed to the next step of completing the square

Step 2: The third term is , ${\left[\left(\frac{1}{2}\right) \left(- 3\right)\right]}^{2} = \frac{9}{4}$

${x}^{2} - 3 x + {\left[\left(\frac{1}{2}\right) \left(- 3\right)\right]}^{2} = 18 + {\left[\left(\frac{1}{2}\right) \left(- 3\right)\right]}^{2}$

${x}^{2} - 3 x + \textcolor{red}{\frac{9}{4}} = 18 + \textcolor{red}{\frac{9}{4}}$

Step 3: Rewrite as a perfect trinomial on the left and simplify the right side of the equation

${x}^{2} - 3 x + {\left(\frac{3}{2}\right)}^{2} = \frac{18}{1} + \frac{9}{4}$

${\left(x - \frac{3}{2}\right)}^{2} = \frac{72}{4} + \frac{9}{4}$

${\left(x - \frac{3}{2}\right)}^{2} = \frac{81}{4}$

Step 4: Solve the equation by taking the square root of both side
remember $\sqrt{{x}^{2}} = \pm x$

$\sqrt{{\left(x - \frac{3}{2}\right)}^{2}} = \sqrt{\frac{81}{4}}$

$x - \frac{3}{2} = \pm \frac{9}{2}$

Step 5: Then solve for $x$

$x = - \frac{9}{2} + \frac{3}{2}$ $\text{ " or }$ $x = \frac{9}{2} + \frac{3}{2}$

So $\text{ } x = - \frac{6}{2} = - 3$

or $\text{ } x = \frac{12}{2} = 6$