# Question #8ce4f

Mar 30, 2017

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

#### Explanation:

Note that
$\textcolor{w h i t e}{\text{XXX}} \textcolor{\mathmr{and} a n \ge}{{\left(x - 1\right)}^{2}} = \textcolor{\mathmr{and} a n \ge}{{x}^{2} - 2 x + 1}$

So
$\textcolor{\mathmr{and} a n \ge}{{\left(x - 1\right)}^{2}} \textcolor{m a \ge n t a}{- 3 x + 4} = 0$
is equivalent to
$\textcolor{\mathmr{and} a n \ge}{{x}^{2} - 2 x + 1} \textcolor{m a \ge n t a}{- 3 x + 4} = 0$
or
${x}^{2} - 5 x + 5 = 0$
or
$\textcolor{b l u e}{1} {x}^{2} \textcolor{red}{- 5} x + \textcolor{g r e e n}{5} = 0$

$\textcolor{w h i t e}{\text{XXX}}$a quadratic of the form $\textcolor{b l u e}{a} {x}^{2} + \textcolor{red}{b} x + \textcolor{g r e e n}{c} = 0$
$\textcolor{w h i t e}{\text{XXX}}$has solutions $x = \frac{- \textcolor{red}{b} \pm \sqrt{\textcolor{red}{b} - 4 \textcolor{b l u e}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{b l u e}{a}}$

So $\textcolor{b l u e}{1} {x}^{2} \textcolor{red}{- 5} x + \textcolor{g r e e n}{5} = 0$

has solutions $x = \frac{+ \textcolor{red}{5} \pm \sqrt{{\left(\textcolor{red}{- 5}\right)}^{2} - 4 \cdot \textcolor{b l u e}{1} \cdot \textcolor{g r e e n}{5}}}{2 \cdot \textcolor{b l u e}{1}}$

$\textcolor{w h i t e}{\text{XXXXXXXX}} = \frac{5 \pm \sqrt{5}}{2}$

Mar 30, 2017

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

#### Explanation:

Let x-1=y so that the given eq becomes ${y}^{2} - 3 y - 3 + 4 = 0$

Or, ${y}^{2} - 3 y + 1 = 0$

Using the quadratic formula, it would be y=$\frac{3 \pm \sqrt{9 - 4}}{2}$

Or, $y = \frac{3}{2} \pm \frac{\sqrt{5}}{2}$

So that $x = 1 + \frac{3}{2} \pm \frac{\sqrt{5}}{2} = \frac{5}{2} \pm \frac{\sqrt{5}}{2}$