# How do you solve using the completing the square method x^2 - 5x + 1 = 0?

May 30, 2016

${x}^{2} - 5 x + 1 = \left(x - \frac{5}{2} - \frac{\sqrt{21}}{2}\right) \left(x - \frac{5}{2} + \frac{\sqrt{21}}{2}\right)$

#### Explanation:

The completing square method is a way of solving quadratic polynomials by reducing then to the structure

${\left(x + \alpha\right)}^{2} - {\beta}^{2} \equiv \left(x + \alpha + \beta\right) \left(x + \alpha - \beta\right)$

The right one is the factorized form.

Now giving a polynomial ${p}_{2} \left(x\right) = {x}^{2} + {a}_{1} x + {a}_{0}$ the reduction begins by making

${x}^{2} + {a}_{1} x + {a}_{0} = {\left(x + \alpha\right)}^{2} - {\beta}^{2}$.

That furnishes

$\left\{\begin{matrix}{a}_{1} = 2 \alpha \\ {a}_{0} = {\alpha}^{2} - {\beta}^{2}\end{matrix}\right.$

Taking $\alpha = {a}_{1} / 2$ we obtain $\beta = \pm \sqrt{{\left({a}_{1} / 2\right)}^{2} - {a}_{0}}$ so the result is

${x}^{2} + {a}_{1} x + {a}_{0} \equiv {\left(x + {a}_{1} / 2\right)}^{2} - {\left(\sqrt{{\left({a}_{1} / 2\right)}^{2}} - {a}_{0}\right)}^{2}$

If ${\left({a}_{1} / 2\right)}^{2} - {a}_{0} < 0$ the roots are complex conjugate.
In the present case we have

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

${x}^{2} - 5 x + 1 = \left(x - \frac{5}{2} - \frac{\sqrt{21}}{2}\right) \left(x - \frac{5}{2} + \frac{\sqrt{21}}{2}\right)$