# How do you solve 18u^2-3u=1?

Jun 28, 2015

$u = \frac{1}{3}$ or $u = - \frac{1}{6}$

#### Explanation:

Method 1
Solutions for $18 {u}^{2} - 3 u = 1$ or equivalently $18 {u}^{2} - 3 u - 1 = 0$
can be determined using the quadratic formula (see bottom, if you don't know this formula).

$u = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(18\right) \left(- 1\right)}}{2 \left(18\right)}$

$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{3 \pm \sqrt{81}}{36}$

$u = \frac{12}{36} = \frac{1}{3}$
or
$u = - \frac{6}{36} = - \frac{1}{6}$

Method 2
If we recognize that (after moving the 1 to the left side) we can factor:
$\textcolor{w h i t e}{\text{XXXX}}$$18 {y}^{2} - 3 u - 1 = 0$
to get
$\textcolor{w h i t e}{\text{XXXX}}$$\left(3 u - 1\right) \left(6 u + 1\right) = 0$

Then either
$\textcolor{w h i t e}{\text{XXXX}}$$\left(3 u - 1\right) = 0$ which implies $u = \frac{1}{3}$
or
$\textcolor{w h i t e}{\text{XXXX}}$$\left(6 u + 1\right) = 0$ which implies $u = - \frac{1}{6}$

Jun 28, 2015

Factor $y = 18 {x}^{2} - 3 x - 1$

#### Explanation:

$y = 18 {x}^{2} - 3 x - 1 = 18 \left(x - p\right) \left(x - q\right) .$
I use the new AC Method (Google, Yahoo Search)
Converted $y ' = {x}^{2} - 3 x - 18$ = (x - p')(x - q').
Factor pairs of (-18)--> (-2, 9)(-3, 6). This sum is 3 = -b
p' = 3 and q' = - 6
$p = \frac{3}{18} = \frac{1}{6}$ and $q = - \frac{6}{18} = - \frac{1}{3}$

Factored form: y = 18(x + 1/6)(x - 1/3) =$\left(6 x + 1\right) \left(3 x - 1\right)$