# How do you solve x^2 - x - 1 = 0?

Oct 4, 2015

Use the quadratic formula or complete the square to find:

$x = \frac{1 \pm \sqrt{5}}{2}$

#### Explanation:

${x}^{2} - x - 1$ is of the form $a {x}^{2} + b x + c$, with $a = 1$, $b = - 1$ and $c = - 1$.

This has solutions given by the quadratic formula

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - \left(4 \times 1 \times \left(- 1\right)\right)}}{2 \cdot 1}$

$= \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$

Method 2 - Completing the square

$0 = {x}^{2} - x - 1 = {\left(x - \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} - 1$

$= {\left(x - \frac{1}{2}\right)}^{2} - \frac{5}{4}$

Add $\frac{5}{4}$ to both ends to get:

${\left(x - \frac{1}{2}\right)}^{2} = \frac{5}{4}$

So:

$x - \frac{1}{2} = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{\sqrt{4}} = \pm \frac{\sqrt{5}}{2}$

Add $\frac{1}{2}$ to both sides to get:

$x = \frac{1}{2} \pm \frac{\sqrt{5}}{2} = \frac{1 \pm \sqrt{5}}{2}$

How are the two methods related?

$0 = a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

Hence:

${\left(x + \frac{b}{2 a}\right)}^{2} = \frac{{b}^{2} / \left(4 a\right) - c}{a} = \frac{{b}^{2} - 4 a c}{4 {a}^{2}}$

So:

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

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

Subtract $\frac{b}{2 a}$ from both sides to get:

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