# How do you solve x^2+3x-9=0 by completing the square?

May 25, 2016

#### Answer:

$x = - \frac{3}{2} \pm \frac{3 \sqrt{5}}{2}$

$x \approx 1.854 \text{ "and" } - 4.854$

#### Explanation:

Not that standard form is $y = a {x}^{2} + b x + c$

and that vertex form is $y = \textcolor{red}{a} {\left(x + \textcolor{red}{\frac{b}{2 a}}\right)}^{2} + c + k$

where $k$ is a correction constant in your case $a = 1$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:$\text{ } y = {x}^{2} + 3 x - 9 = 0$

write as: $0 = \left({x}^{2} + 3 x\right) - 9$

Add the correction constant $k$

$0 = \left({x}^{2} + 3 x\right) - 9 + k$

Remove the $x$ from $3 x$

$0 = \left({x}^{2} + 3\right) - 9 + k$

Halve the 3

$0 = \left({x}^{2} + \frac{3}{2}\right) - 9 + k$

Move the power from ${x}^{2}$ to outside the bracket

0=(x+3/2)^2-9+k" "color(red)(larr "the error comes from "a(3/(2a))^2)

but $k + {\left(\frac{3}{2}\right)}^{2} = 0 \implies k = - \frac{9}{4}$ giving

color(blue)("Vertex form "->0=(x+3/2)^2- 45/4
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Take the $- \frac{45}{4}$ to the other side of the equals and change its sign.

${\left(x + \frac{3}{2}\right)}^{2} = + \frac{45}{4}$

Square root both sides

$x + \frac{3}{2} = \sqrt{\frac{45}{4}}$

Move the $\frac{3}{2}$ to the other side of =

$x = - \frac{3}{2} \pm \frac{\sqrt{5 \times 9}}{\sqrt{4}}$

$\textcolor{b l u e}{x = - \frac{3}{2} \pm \frac{3 \sqrt{5}}{2}}$

$\textcolor{b l u e}{x \approx 1.854 \text{ "and" } - 4.854}$ 