# How do you solve using the completing the square method x^2 - 14x = 0?

Jul 11, 2016

#### Answer:

$x$ is $0 , 14$

#### Explanation:

We begin with ${x}^{2} - 14 x = 0$. I like to deal with equations in standard form ($\textcolor{red}{a} {\textcolor{g r e e n}{x}}^{2} + \textcolor{b l u e}{b} \textcolor{g r e e n}{x} + \textcolor{\mathmr{and} a n \ge}{c}$), so I'll just rewrite this to ${x}^{2} - 14 x + 0 = 0$. From here, we need to solve for $x$, and we need to do that by completing the square. Completing the square is a method where we take an equation that is not a perfect square and find a value that could make it factorable.

For us, our first step is to make sure that the $\textcolor{red}{a}$ in $\textcolor{red}{a} {\textcolor{g r e e n}{x}}^{2} + \textcolor{b l u e}{b} \textcolor{g r e e n}{x} + \textcolor{\mathmr{and} a n \ge}{c}$ is a $1$. Is that the case for ${x}^{2} - 14 x + 0 = 0$? Yes, it is.

Now we take the second coefficient ($\textcolor{b l u e}{b}$), and divide it in half ($- \frac{14}{2}$), in our case giving us $- 7$. We then take that value and square it, giving us $49$ ($- {7}^{2}$). We take that number and add it to our equation, like this:

${x}^{2} - 14 x + 49 + 0 = 0$.

WAIT!!!! We just added a random number into this equation! We can't do that. We need to keep the equation equal. We could add a $49$ on the other side, or we could just subtract the $49$. Then the numbers cancel each other out and don't change the value of the equation.

NOW we have ${x}^{2} - 14 x + 49 - 49 + 0 = 0$. Now, we went to a lot of effort to get ${x}^{2} - 14 x + 49$, because this is a perfect square. We can rewrite that part of the equation into ${\left(x - 7\right)}^{2}$. We still need to deal with $- 49 + 0$, so we just combine them to give us $- 49$. Put it together and we get ${\left(x - 7\right)}^{2} - 49 = 0$.

Now we just need to solve for $x$. I'll go through this part quickly.

${\left(x - 7\right)}^{2} - 49 = 0$
${\left(x - 7\right)}^{2} = 49$
$\sqrt{{\left(x - 7\right)}^{2}} = \sqrt{49}$
$x - 7 = \pm 7$
$x = 7 \pm 7$
$x = 0 , 14$