# How do you solve x^2-20x+99=0?

Apr 10, 2017

See the solution process below:

#### Explanation:

Because $- 11 - 9 = - 20$ and $- 9 \times - 11 = 99$ we can factor this equation as:

$\left(x - 9\right) \left(x - 11\right) = 0$

We can now solve each term for $0$ to find the solutions:

Solution 1)

$x - 9 = 0$

$x - 9 + \textcolor{red}{9} = 0 + \textcolor{red}{9}$

$x - 0 = 9$

$x = 9$

Solution 2)

$x - 11 = 0$

$x - 11 + \textcolor{red}{11} = 0 + \textcolor{red}{11}$

$x - 0 = 11$

$x = 11$

The solution is: $x = 9$ and $x = 11$

Apr 10, 2017

$x = 9 , 11$

#### Explanation:

There are a number of ways to solve polynomials. We can graph the equation, or use the quadratic formula, or factor. Let's start with factoring and see if we can solve it that way. Then we'll move on to the quadratic formula

${x}^{2} - 20 x + 99$
We are looking for two numbers that combine to $- 20$ and multiply to $99$.

$\textcolor{w h i t e}{. .} - 20$
$\textcolor{w h i t e}{. .}$x$\textcolor{w h i t e}{.} 99$
..............
$\textcolor{w h i t e}{. .} 1 \cdot 99$
$\textcolor{w h i t e}{. .} 3 \cdot 33$
$\textcolor{w h i t e}{. .} \textcolor{\mathmr{and} a n \ge}{9 \cdot 11}$

$- 9 + - 11$ gives us $- 20$, and $- 9 \cdot - 11$ equals $99$. So our factors are $\left(x - 9\right)$ and $\left(x - 11\right)$.

$\left(x - 9\right) \left(x - 11\right) = 0$
Now we set each factor equal to $0$

Factor 1
$x - 9 = 0$
$x = 9$

Factor 2
$x - 11 = 0$
$x = 11$

So, we solved the equation without using quadratic formula. If you want to use the quadratic formula, just plug $a {x}^{2} + b x + c$ into $\frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 a}$. Both methods will give you the same answer.

Jun 30, 2017

9 and 11

#### Explanation:

${x}^{2} - 20 x + 99 = 0$
Find 2 real roots knowing sum (-b = 20) and product (c = 99).
They are: 9 and 11. (because sum = 20 and product 99)

NOTE: When a = 1, there is no need to solving by grouping and solving the 2 binomials.