# How do you describe the nature of the roots of the equation x^2-4x+4=0?

Oct 17, 2017

We can factor the expression on the right as:

$\left(x - 2\right) \left(x - 2\right) = 0$ or ${\left(x - 2\right)}^{2} = 0$

Therefore the two roots for this quadratic are the same: $x = 2$

#### Explanation:

We can also use the discriminate to show this same result:

For $a {x}^{2} + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:

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

The discriminate is the portion of the quadratic equation within the radical: ${\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}$

If the discriminate is:
- Positive, you will get two real solutions
- Zero you get just ONE solution
- Negative you get complex solutions

To find the discriminant for this problem substitute:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 4}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{4}$ for $\textcolor{g r e e n}{c}$

${\textcolor{b l u e}{\left(- 4\right)}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{4}\right) \implies$

$16 - 16 \implies$

$0$

Because the discriminate is $0$ you get just ONE solution.