# Solve the equation. Leave answer in simplest radical form. n^2 - 2 n + 1 = 5.

## Question:

{eq}n^2 - 2 n + 1 = 5 {/eq}.

An equation which is of the form {eq}an^2+bn+c=0 {/eq} where {eq}a, b \text{ and } c {/eq} are constants is called a quadratic equation in {eq}x {/eq}. It has two (either same or distinct) roots. The roots can be solved by using the quadratic formula which states:

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

The given equation is:

$$n^2 - 2 n + 1 = 5 \\ \text{Subtracting 5 from both sides}, \\ n^2-2n-4=0$$

Comparing it with {eq}an^2+bn+c=0 {/eq}, we get:

$$a=1 \\ b=-2 \\ c=-4$$

Substitute all these values in the quadratic formula:

\begin{align} n &=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\[0.4cm] &= \frac{-(-2) \pm \sqrt{(-2)^{2}-4 \cdot 1(-4)}}{2 \cdot 1} \\[0.4cm] & = \dfrac{2 \pm \sqrt{4+16}}{2} \\[0.4cm] &= \dfrac{2 \pm \sqrt{20}}{2} \\[0.4cm] &= \dfrac{2 \pm 2 \sqrt{5}}{2} & [\because \sqrt{20} = \sqrt{4} \times \sqrt{5}= 2 \sqrt{5} ] \\ &= 1 \pm \sqrt{5} \end{align}

Therefore, {eq}\boxed{\mathbf{n=1+\sqrt{5}; \,\,\, n=1-\sqrt{5}}} {/eq}.