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

Jul 24, 2017

$x = \pm \sqrt{\frac{10}{9}} + \frac{1}{3}$, or $\approx 1.39$ and $\approx - 0.72$

#### Explanation:

First things first, in order to complete the square, the leading coefficient ($\textcolor{red}{3} {x}^{2} - 2 x - 3$) must be $1$. To do that, we need to factor out a $3$ from the equation.

$3 {x}^{2} - 2 x - 3 = 0$

$3 \left({x}^{2} - \frac{2}{3} x - 1\right) = 0$

Now we have the beginning. Completing the square can look scary, but it's really just a process, and if you understand the steps, it becomes pretty simple.

The first step is getting the leading coefficient to $1$. After that, we need to take the middle term, $- \frac{2}{3}$ and "do some stuff with it" (you'll see in a minute)

So, we'll take $\left(- \frac{2}{3}\right)$ and divide it by $2$, which gives us $- \frac{1}{3}$. Now we square the solution, which equals $\frac{1}{9}$.

We did all of this because we needed to find the value that will make the left side of our equation, ${x}^{2} - \frac{2}{3} x - 1 = 0$, a perfect square, which is $\frac{1}{9}$.

Now that we have our missing value, we need to add it to our equation.

${x}^{2} - \frac{2}{3} x - 1 + \textcolor{red}{\frac{1}{9}} \text{ } = 0$

But wait!! We can't just add a random number into an equation! An equation is all about balance (the root word is equal). You can't just introduce a new value. But.... if you add $500$, and then immediately subtract $500$, the final result is $0$.

So, if we add $\textcolor{red}{\frac{1}{9}}$, and then subtract $\textcolor{red}{\frac{1}{9}}$, then technically we haven't changed anything

${x}^{2} - \frac{2}{3} x - 1 + \textcolor{red}{\frac{1}{9}} + \textcolor{red}{- \frac{1}{9}} = 0$

Let's re-order this:

$\textcolor{g r e e n}{{x}^{2} - \frac{2}{3} x + \frac{1}{9}} \textcolor{b l u e}{- 1 - \frac{1}{9}} = 0$

$\textcolor{g r e e n}{{x}^{2} - \frac{2}{3} x + \frac{1}{9}}$ is a perfect square (that was the whole point of all of this, after all). Let's factorise it.

$\textcolor{g r e e n}{{x}^{2} - \frac{2}{3} x + \frac{1}{9}} = \textcolor{g r e e n}{{\left(x - \frac{1}{3}\right)}^{2}}$

Let's simplify this: $\textcolor{b l u e}{- 1 - \frac{1}{9}}$ equals $\textcolor{b l u e}{- \frac{10}{9}}$

So, now we have $\textcolor{g r e e n}{{\left(x - \frac{1}{3}\right)}^{2}} \textcolor{b l u e}{- \frac{10}{9}} = 0$

$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$

Now, let's solve this puppy!

${\left(x - \frac{1}{3}\right)}^{2} - \frac{10}{9} = 0$

add $\frac{10}{9}$ to both sides

${\left(x - \frac{1}{3}\right)}^{2} = \frac{10}{9}$

take a square root of both sides

$\sqrt{{\left(x - \frac{1}{3}\right)}^{2}} = \pm \sqrt{\frac{10}{9}}$

$x - \frac{1}{3} = \pm \sqrt{\frac{10}{9}}$

add $\frac{1}{3}$ to both sides

$x = \pm \sqrt{\frac{10}{9}} + \frac{1}{3}$,

or $\approx 1.39$ and $\approx - 0.72$

Jul 24, 2017

$x = \frac{\sqrt{10} + 1}{3}$
$x = \frac{- \left(\sqrt{10}\right) + 1}{3}$

#### Explanation:

Given -

$3 {x}^{2} - 2 x - 3 = 0$

Take the constant to the right

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

Divide both sides by the coefficient of ${x}^{2}$

$\frac{3 {x}^{2}}{3} - \frac{2}{3} x = \frac{3}{3}$

${x}^{2} - \frac{2}{3} x = 1$

Take half of the coefficient of $x$ and square it

Half the coefficient of $x = - \frac{2}{3} \div 2 = - \frac{2}{3} \times \frac{1}{2} = - \frac{2}{6}$

Square of $- \frac{2}{6} = \frac{4}{36} = \frac{1}{9}$

Add $\frac{1}{9}$ to both the sides

${x}^{2} - \frac{2}{3} x + \frac{1}{9} = 1 + \frac{1}{9} = \frac{9 + 1}{9} = \frac{10}{9}$

${\left(x - \frac{1}{3}\right)}^{2} = \frac{10}{9}$

Taking square root on both sides

$x - \frac{1}{3} = \pm \sqrt{\frac{10}{9}} = \pm \frac{\sqrt{10}}{3}$
$x = \pm \frac{\sqrt{10}}{3} + \frac{1}{3} = \pm \frac{\sqrt{10} + 1}{3}$

$x = \frac{\sqrt{10} + 1}{3}$
$x = \frac{- \left(\sqrt{10}\right) + 1}{3}$