# How do you solve using the quadratic formula for y = x^2 - 4x + 4?

##### 1 Answer
May 14, 2015

The answer is : $y = {x}^{2} - 4 x + 4 = {\left(x - 2\right)}^{2}$. Your function has one zero, with $x = 2$.

The usual form of a quadratic function is : $y = a {x}^{2} + b x + c$

Your function is $y = 1 {x}^{2} - 4 x + 4$.

Therefore $a = 1$, $b = - 4$ and $c = 4$

The quadratic formula gives you the values of $x$, with which your $y = 0$.

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

Since we have a square root, $\Delta \ge 0$. If not, the function don't have any zeros.

Let's calculate the zeros of your function :

$\Delta = {\left(- 4\right)}^{2} - 4 \cdot 1 \cdot 4 = 16 - 16 = 0$

${x}_{1} = \frac{4 + \sqrt{0}}{2} = {x}_{2} = \frac{4 - \sqrt{0}}{2} =$2

Therefore, $y = {x}^{2} - 4 x + 4 = {\left(x - 2\right)}^{2}$. Your function has one zero, with $x = 2$.