# How do you solve x^2+3x+21=22 by completing the square?

Mar 28, 2017

$x = - \frac{3}{2} \pm \frac{\sqrt{13}}{2}$

#### Explanation:

Steps:

Subtract 21 from both sides

x^2+3x + ? = 1

In order to find ?, you must divide $3$ by $2$ or $\left(\frac{b}{2}\right)$and square it and add it to both sides:

? = (3/2)^2 =9/4

${x}^{2} + 3 x + \frac{9}{4} = 1 + \frac{9}{4}$

Now we can factor the left side and simplify the right.

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

${\left(x + \frac{3}{2}\right)}^{2} = \frac{13}{4}$

Finally, solve for $x$

$x + \frac{3}{2} = 0 \to x = - \frac{3}{2}$

${x}^{2} - \frac{3}{14} = 0 \to {x}^{2} = \frac{13}{14} \to x = \pm \sqrt{\frac{13}{4}} \to x = \pm \frac{\sqrt{13}}{2}$

Mar 28, 2017

 x = (-3 +- sqrt13)/2

#### Explanation:

Squares of form ${\left(x + a\right)}^{2}$ are expanded like this:

${x}^{2} + 2 a x + {a}^{2}$

Since our equation starts with ${x}^{2} + 3 x \ldots$, we know that $3 = 2 a$, so $\frac{3}{2} = a$. That means that to complete the square, we need the last term to be ${a}^{2} = {\left(\frac{3}{2}\right)}^{2} = \frac{9}{4}$.

To do this, we can subtract $18.75$ from both sides and simplify.

$\textcolor{w h i t e}{\text{X}} {x}^{2} + 3 x + 21 = 22$

$\textcolor{w h i t e}{\text{X}} {x}^{2} + 3 x + \frac{9}{4} = \frac{13}{4}$

$\textcolor{w h i t e}{\text{XX..}} {\left(x + 1.5\right)}^{2} = \frac{13}{4}$

Now we can take the square root of both sides. Remember that the square root can be either positive or negative, since when squared, the negative sign disappears.

color(white)"" x + 3/2 = +- sqrt13/2

$\textcolor{w h i t e}{\text{XXX.}} x = \frac{- 3 \pm \sqrt{13}}{2}$