# How do you solve x^2 - 3x = 18 by completing the square?

Jul 14, 2018

$x = - 3 \text{ or } x = 6$

#### Explanation:

$\text{add "(1/2"coefficient of the x-term")^2" to both sides}$

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

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

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x - \frac{3}{2}\right)}^{2}} = \pm \sqrt{\frac{81}{4}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

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

$\text{add "3/2" to both sides}$

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

$x = \frac{3}{2} - \frac{9}{2} = - 3 \text{ or } x = \frac{3}{2} + \frac{9}{2} = 6$

Jul 14, 2018

$x = 6 \mathmr{and} x = - 3$

#### Explanation:

Here ,

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

$\implies {x}^{2} - 3 x + k = k + 18. . . \to \left(I\right)$

We have to find $k$ such that ,${x}^{2} - 3 x + k$
forms a perfect square ,where

color(blue)(1^(st)term=x^2 ,

color(blue)(2^(nd)term=-3x

color(blue)(3^(rd)term=k

Formula to find ${3}^{r d} t e r m$ is :

$\textcolor{red}{{3}^{r d} t e r m = {\left({2}^{n d} t e r m\right)}^{2} / \left(4 \times {1}^{s t} t e r m\right)} \ldots \to \left(A\right)$

So,

$k = {\left(- 3 x\right)}^{2} / \left(4 \cdot {x}^{2}\right) = \frac{9 {x}^{2}}{4 {x}^{2}} = \frac{9}{4}$

Subst. $k = \frac{9}{4}$ into $\left(I\right)$ ,we get

$\implies {x}^{2} - 3 x + \frac{9}{4} = \frac{9}{4} + 18 = \frac{81}{4}$

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

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

$\implies x - \frac{3}{2} = \frac{9}{2} \mathmr{and} x - \frac{3}{2} = - \frac{9}{2}$

$\implies x = \frac{9}{2} + \frac{3}{2} \mathmr{and} x = \frac{3}{2} - \frac{9}{2}$

$\implies x = \frac{12}{2} \mathmr{and} x = - \frac{6}{2}$

$\implies x = 6 \mathmr{and} x = - 3$

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Note:
We can use Formula $\left(A\right)$ without any doubt.