# How do you solve (c - 3) ( c - 1) = 6?

Aug 1, 2017

$c = 2 \pm \sqrt{7}$

#### Explanation:

First we can put the equation into standard form $a {x}^{2} + b x + c = 0$

$\left(c - 3\right) \left(c - 1\right) = 6$

$\iff$ Use FOIL to expand the polynomial

${c}^{2} - 4 c + 3 = 6$

$\iff$ Subtract $6$ from both sides

${c}^{2} - 4 c - 3 = 0$

Since you can't factor this polynomial, we can use the quadratic formula

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

In this case

$a = 1$

$b = - 4$

$c = - 3$

and we replace $x$ with the variable $c$

then we plug in

$\implies c = \frac{4 \pm \sqrt{16 - 4 \left(1\right) \left(- 3\right)}}{2} = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2}$

Since $\sqrt{a b} = \sqrt{a} \sqrt{b}$

$= \frac{4 \pm \sqrt{4} \sqrt{7}}{2} = \frac{4 \pm 2 \sqrt{7}}{2} = 2 \pm \sqrt{7}$

Which means that

$c = 2 + \sqrt{7} \mathmr{and} c = 2 - \sqrt{7}$

Aug 1, 2017

Complete the square to find:

$c = 2 + \sqrt{7} \text{ }$ or $\text{ } c = 2 - \sqrt{7}$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We can use this with $a = c - 2$ and $b = \sqrt{7}$ as follows.

Given:

$\left(c - 3\right) \left(c - 1\right) = 6$

Multiply out the left hand side to get:

${c}^{2} - 4 c + 3 = 6$

Subtract $6$ from both sides to get:

${c}^{2} - 4 c - 3 = 0$

Complete the square and use the difference of squares identity to find:

$0 = {c}^{2} - 4 c - 3$

$\textcolor{w h i t e}{0} = {c}^{2} - 4 c + 4 - 7$

$\textcolor{w h i t e}{0} = {\left(c - 2\right)}^{2} - {\left(\sqrt{7}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(c - 2\right) - \sqrt{7}\right) \left(\left(c - 2\right) + \sqrt{7}\right)$

$\textcolor{w h i t e}{0} = \left(c - 2 - \sqrt{7}\right) \left(c - 2 + \sqrt{7}\right)$

Hence:

$c = 2 + \sqrt{7} \text{ }$ or $\text{ } c = 2 - \sqrt{7}$