# How do you solve x^2 - 5x = 36?

May 1, 2018

$x = - 4 , 9$

#### Explanation:

Subtract 36 from both sides

${x}^{2} - 5 x - 36 = 0$

This is the equation of a quadratic so wee need the x-intercepts.

Two numbers that multiply to 36 are 4 and -9, and they add to -5 so your equation can be

$\left(x + 4\right) \left(x - 9\right) = 0$

so $x = - 4 , 9$

May 1, 2018

${x}^{2} - 5 x = 36$
minus 36 (both sides)
${x}^{2} - 5 x - 36 = 0$
Now, we need to find two numbers (for exampe a and b).
$a \cdot b = - 36$ and $a + b = - 5$
$a = - 9$
$b = + 4$
So, the solution is: $\left(x - 9\right) \cdot \left(x + 4\right) = 0$
We want these brackets to be equal to zero.
$x = 9$
$x = - 4$
Two solutions.

May 1, 2018

Solve by either factoring or using The Quadratic Formula to find $x = \left\{- 4 , 9\right\}$

#### Explanation:

The first step is to set the stated expression equal to zero. we will do this by subtracting 36 from both sides:

${x}^{2} - 5 x \textcolor{red}{- 36} = \cancel{36 \textcolor{red}{- 36}}$

${x}^{2} - 5 x - 36 = 0$

Now, we can either factor this equation OR use The Quadratic Formula to find the values of $x$ that satisfy the equation. Let's start with factoring.

Based on the order of the equation (second order due to the presence of ${x}^{2}$), we will assume this will have two factors. Additionally, since the third factor is negative, we know that one factor is positive and the other is negative:

$\left(x + a\right) \left(x + b\right) = {x}^{2} + \left(a + b\right) x + a b$

Now, we just need to figure out what two numbers satisfy the following system:

$a + b = - 5$
$a b = - 36$

We know that 36 is a square, so a possible combination is 6 and -6. However, that would not satisfy the first equation. Other factors of -36 include:

$\left\{\pm 36 , \pm 1\right\}$
$\left\{\pm 18 , \pm 2\right\}$
$\left\{\pm 12 , \pm 3\right\}$
$\left\{\pm 9 , \pm 4\right\}$

The only set there that would satisfy the first equation though, is $\left\{- 9 , 4\right\}$. Now that we have our factors, we can write the factored equation and evaluate:

${x}^{2} - 5 x - 36 = \left(x + 4\right) \left(x - 9\right) = 0$

When $x = - 4$:

$\left(\textcolor{p u r p \le}{- 4} + 4\right) \left(\textcolor{p u r p \le}{- 4} - 9\right) = 0$

$\left(0\right) \left(- 13\right) = 0 \Rightarrow 0 = 0$

When $x = 9$:

$\left(\textcolor{p u r p \le}{9} + 4\right) \left(\textcolor{p u r p \le}{9} - 9\right) = 0$

$\left(13\right) \left(0\right) = 0 \Rightarrow 0 = 0$

We have our two solutions for $x$:

$\textcolor{g r e e n}{x = \left\{- 4 , 9\right\}}$

We plug in our equations coefficients into the formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{red}{a = 1}$
$\textcolor{b l u e}{b = - 5}$
$\textcolor{p u r p \le}{c = - 36}$

$x = \frac{- \textcolor{b l u e}{\left(- 5\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 5\right)}}^{2} - 4 \textcolor{red}{\left(1\right)} \textcolor{p u r p \le}{\left(- 36\right)}}}{2 \textcolor{red}{\left(1\right)}}$

$x = \frac{5 \pm \sqrt{25 - 4 \left(- 36\right)}}{2}$

$x = \frac{5 \pm \sqrt{25 + 144}}{2}$

$x = \frac{5 \pm \sqrt{169}}{2}$

$x = \frac{5 \pm 13}{2}$

$x = \left\{\frac{5 - 13}{2} , \frac{5 + 13}{2}\right\}$

$x = \left\{- \frac{8}{2} , \frac{18}{2}\right\}$

$\textcolor{g r e e n}{x = \left\{- 4 , 9\right\}}$