# How do you solve 100x^2-800x+1500=0?

Dec 12, 2016

$x = 5$ and $x = 3$

#### Explanation:

First, divide both sides of the equation by $100$ to make this simpler to work with:

$\frac{1}{100} \left(100 {x}^{2} - 800 x + 1500\right) = \frac{0}{100}$

$\frac{100 {x}^{2}}{100} - \frac{800 x}{100} + \frac{1500}{100} = 0$

${x}^{2} - 8 x + 15 = 0$

Now we can play with multipliers of 15 (1x15, 3x5, 5x3, 15x1) to factor the quadratic equation:

$\left(x - 5\right) \left(x - 3\right) = 0$

We can now solve each term for $0$:

$x - 5 = 0$

$x - 5 + 5 = 0 + 5$

$x - 0 = 5$

$x = 5$

and

$x - 3 = 0$

$x - 3 + 3 = 0 + 3$

$x - 0 = 3$

$x = 3$

Dec 12, 2016

$\textcolor{g r e e n}{x = 3}$ or $\textcolor{g r e e n}{x = 5}$

#### Explanation:

If
$\textcolor{w h i t e}{\text{XXX}} 100 {x}^{2} - 800 x + 1500 = 0$
then (after dividing both sides by $100$)
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 8 + 15 = 0$

We would like to factor this in the form:
$\textcolor{w h i t e}{\text{XXX}} \left(x - a\right) \left(x - b\right) = 0$
and since
$\textcolor{w h i t e}{\text{XXX}} \left(x - a\right) \left(x - b\right) = {x}^{2} + \left(- a - b\right) x + a b$
we are looking for values of $a$ and $b$
such that
$\textcolor{w h i t e}{\text{XXX}} - a - b = - 8 \rightarrow a + b = 8$
and
$\textcolor{w h i t e}{\text{XXX}} a b = 15$

Checking factors of $15$, we quickly find the pair $3$ and $5$ that satisfy our requirement.

So we have
$\textcolor{w h i t e}{\text{XXX}} \left(x - 3\right) \left(x - 5\right) = 0$

From which it follows that
{: ("either ",(x-3)=0," or ",(x-5)=0), (,rarr x=3,,rarrx=5) :}