# How do you solve using the completing the square method 3x^2+10x+13?

##### 1 Answer
Dec 20, 2017

Required Solutions are

color(blue)(x = (-5/3) +- [sqrt(14)/3]i

#### Explanation:

We are given the quadratic expression

color(red)(f(x)=3x^2+10x+13

We must use Completing The Square Method to find the solutions.

Let us assume that $\textcolor{red}{\text{ } 3 {x}^{2} + 10 x + 13 = 0}$

We will find the solutions in steps.

$\textcolor{g r e e n}{S t e p .1}$

In this step,

we will move the constant term to Right-Hand Side (RHS)

Hence, we get

$\textcolor{red}{\text{ } 3 {x}^{2} + 10 x = - 13}$

$\textcolor{g r e e n}{S t e p .2}$

Divide each term by 3 to get the coefficient of the ${x}^{2}$ term as $1$

$\left(\frac{3}{3}\right) {x}^{2} + \left(\frac{10}{3}\right) x = \left(- \frac{13}{3}\right)$

$\Rightarrow {x}^{2} + \left(\frac{10}{3}\right) x = \left(- \frac{13}{3}\right)$

$\textcolor{g r e e n}{S t e p .3}$

We will add a value to each side

$\Rightarrow {x}^{2} + \left(\frac{10}{3}\right) x + \textcolor{red}{\square} = - \left(\frac{13}{3}\right) + \textcolor{red}{\square}$

In the next step, we need to figure out how-to find the value that goes into the $\textcolor{red}{R E D}$ box

$\textcolor{g r e e n}{S t e p .4}$

Divide the coefficient of the $x$ term by $2$ and square it.

The result of the calculation will replace the $\textcolor{red}{R E D}$ box in the next step.

The calculation is done as follows:

Coefficient of the x-term is $\left(\frac{10}{3}\right)$

If we divide by 2, we will get $\left(\frac{10}{6}\right)$

When we square this intermediate result, we get

${\left(\frac{10}{6}\right)}^{2} \Rightarrow {\left(\frac{5}{3}\right)}^{2} \Rightarrow \left(\frac{25}{9}\right)$

Hence, the value $\left(\frac{25}{9}\right)$ will replace the $\textcolor{red}{R E D}$ box in the next step.

$\textcolor{g r e e n}{S t e p .5}$

$\Rightarrow {x}^{2} + \left(\frac{10}{3}\right) x + \left(\frac{25}{9}\right) = - \left(\frac{13}{3}\right) + \left(\frac{25}{9}\right)$

We can write the Left-Hand-Side (LHS) as

$\Rightarrow {\left(x + \textcolor{red}{\frac{5}{3}}\right)}^{2} = \frac{- 39 + 25}{9}$

Note that we get the value $\textcolor{red}{\frac{5}{3}}$ by dividing the coefficient of the x-term by $2$

$\textcolor{g r e e n}{S t e p .6}$

We will now work on

${\left(x + \textcolor{red}{\frac{5}{3}}\right)}^{2} = \frac{- 39 + 25}{9}$ **

We will take Square Root on both sides to simplify:

Hence, we get

$\sqrt{x + {\left(\frac{5}{3}\right)}^{2}} = \pm \sqrt{\frac{- 39 + 25}{9}}$

Square root and Square cancel out.

$\therefore$ on simplification we get,

$x + \frac{5}{3} = \pm \sqrt{\frac{- 14}{9}}$

$\Rightarrow x + \frac{5}{3} = \pm \frac{\sqrt{14 \cdot \left(- 1\right)}}{\sqrt{9}}$

Note that, in complex number system,

$i = \sqrt{- 1}$

${i}^{2} = i \cdot i = \left(- 1\right)$

Hence,

$\Rightarrow x + \frac{5}{3} = \pm \frac{\sqrt{14 \cdot {i}^{2}}}{3}$

$x = - \frac{5}{3} \pm \left[\frac{\sqrt{14}}{3}\right] i$

Hence, required solutions are

color(blue)(x = (-5/3) +- [sqrt(14)/3]i

Hope this helps.