How do I go from $\frac{33}{64} - 2x + x^2 \longrightarrow\bigg(x - \frac{2 + \sqrt{4 +33/16}}{2}\bigg)\bigg(x - \frac{2 - \sqrt{4 +33/16}}{2}\bigg)$ and then to $\big(x - 1-\frac{\sqrt{33}}{8}\big)\big(x - 1 + \frac{\sqrt{31}}{8}\big)?$
For the first, I know that it has something with the quadratic equation to do, but what is that? And how do I go from the second to the third?
$$ \begin{align} &\frac{33}{64} - 2x + x^2 \\ &= x^2 - 2x + \frac{33}{64} \\ &= x^2 - 2x + 1 - 1 + \frac{33}{64} \\ &= (x-1)^2 - 31/64 \\ &= (x-1)^2 - 31/64 \\ &= (x-1)^2 - 31/64 \\ &= \left( (x-1) - \sqrt{31}/8 \right)\cdot\left( (x-1) + \sqrt{31}/8 \right) \\ &= \color{red}{\left( x-1 - \sqrt{31}/8 \right)\cdot\left( x-1 + \sqrt{31}/8 \right)} \\ &= \left( x-\frac {2 + \sqrt{31}/4}2 \right)\cdot\left( x-\frac{2 - \sqrt{31}/4}{2} \right) \\ &= \left( x-\frac {2 + \sqrt{31/16}}2 \right)\cdot\left( x-\frac{2 - \sqrt{31/16}}{2} \right) \\ & = \left(x - \frac{2 + \sqrt{4 \color{red}-33/16}}{2}\right)\cdot\left(x - \frac{2 - \sqrt{4 \color{red}-33/16}}{2}\right) \end{align} $$
Please note the minus inside square roots!
The quadratic eqation:
If $ax^2 +bx +c =0$ then
$x^2 +\frac {b }{a}x =-c/a$
$x^2 + 2\frac {b}{2a}x +\frac {b}{2a}^2 = -c/a +\frac {b}{2a}^2=\frac {b^2 - 4ac}{4a^2} $
$(x+b/2a)^2 = \frac {b^2 - 4ac}{4a^2} $
$x+ b/2a = \pm \frac {\sqrt { b^2 - 4ac }}{2a} $
$x=\frac {-b\pm \sqrt { b^2 - 4ac } }{2a}$
And therefore:
$ax^2 + bx+ c= a (x + \frac {b +\sqrt { b^2 - 4ac } }{2a} )(x + \frac {b- \sqrt { b^2 - 4ac } }{2a})$
So if $ax^2 + bx + c >0; a >0$ then $ (x + \frac {b +\sqrt { b^2 - 4ac } }{2a} )(x + \frac {b- \sqrt { b^2 - 4ac } }{2a}) > 0$ and either both terms are positive or both terms are negative.
The final bit is just arithmetic on fractions.
There is an error. It should be $1 \pm \sqrt {97}/8$
