# What is the quadratic formula for f(b)=b^2 - 4b + 4 = 0?

Oct 22, 2015

Rewriting $f \left(b\right)$ as $f \left(x\right)$ will permit you to use the standard formula with less confusion (since the standard quadratic formula uses $b$ as one of its constants)

#### Explanation:

(since the given equation uses $b$ as a variable, we will need to express the quadratic formula, which normally uses $b$ as a constant, with some variant, $\hat{b}$.

To help reduce confusion, I will rewrite the given $f \left(b\right)$as
$\textcolor{w h i t e}{\text{XX}} f \left(x\right) = {x}^{2} - 4 x + 4 = 0$

$\textcolor{w h i t e}{\text{XX}} \hat{a} {x}^{2} + \hat{b} x + \hat{c} = 0$
the solution given by the quadratic equation is
$\textcolor{w h i t e}{\text{XX}} x = \frac{- \hat{b} \pm \sqrt{{\hat{b}}^{2} - 4 \hat{a} \hat{c}}}{2 \hat{a}}$

With $\hat{a} = 1$, $\hat{b} = - 4$, and $\hat{c} = + 4$
we get
$\textcolor{w h i t e}{\text{XX}} b = \left(x =\right) \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} + 4 \left(1\right) \left(4\right)}}{2 \left(1\right)}$