# How do you solve \frac { 1} { c + 10} = \frac { c } { 26}?

May 18, 2018

${c}_{1 , 2} = - 5 \setminus \pm \setminus \sqrt{51}$

#### Explanation:

Bring everything to the $L H S$:

$\setminus \frac{1}{c + 10} - \setminus \frac{c}{26} = 0$

Least common denominator is $26 \left(c + 10\right)$:

$\setminus \frac{26 - c \left(c + 10\right)}{26 \left(c + 10\right)} = \setminus \frac{- {c}^{2} - 10 c + 26}{26 \left(c + 10\right)} = 0$

A fraction equals zero only if its numerator equals zero:

$- {c}^{2} - 10 c + 26 = 0 \setminus \iff {c}^{2} + 10 c - 26 = 0$

\frac{-10 \pm \sqrt(100 + 104)}{2} = \frac{-10 \pm sqrt(4*51)}{2} = \frac{cancel(-10)^{-5} \pm cancel(2)sqrt(51)}{cancel(2)}

So, the two solutions are

${c}_{1 , 2} = - 5 \setminus \pm \setminus \sqrt{51}$

May 18, 2018

A slightly different beginning.

$c = 5 \pm \sqrt{51}$

#### Explanation:

Given: $\frac{1}{c + 10} = \frac{c}{26}$

I wish to have the $c + 10$ as a numerator so turn everything upside down.

$\frac{c + 10}{1} = \frac{26}{c}$

Multiply both sides by c

${c}^{2} + 10 c = 26$

Subtract 26 from both sides

${c}^{2} + 10 x - 26 = 0 \leftarrow \text{ A quadratic}$

You will not have whole number factors so use the formula.

In its normally remembered form we have:
$a {x}^{2} + b x + c = 0 \to x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

a=1; b=10; c=-26 and x->c

$c = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \left(1\right) \left(- 26\right)}}{2 \left(1\right)}$

$c = - 5 \pm \frac{\sqrt{100 + 104}}{2}$

You are looking for squared factors that you can 'take outside' the root.

If you are ever uncertain about factoring larger values draw a quick sketch of a prime factor tree. $c = - 5 \pm \frac{\sqrt{{2}^{2} \times 51}}{2}$

$c = 5 \pm \frac{\cancel{2} \sqrt{51}}{\cancel{2}}$

$c = 5 \pm \sqrt{51}$