Question

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

Answer 1

`c_{1,2} = -5 \pm \sqrt(51)`

Explanation:

Bring everything to the `LHS`:

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

Least common denominator is `26(c+10)`:

`\frac{26 - c(c+10)}{26(c+10)} = \frac{-c^2-10c+26}{26(c+10)} = 0`

A fraction equals zero only if its numerator equals zero:

`-c^2-10c+26=0 \iff c^2+10c-26=0`

Quadratic formula to solve:

#\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 \pm \sqrt(51)`

Answer 2

A slightly different beginning.

`c=5+-sqrt(51)`

Explanation:

Given: `1/(c+10)=c/26`

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

`(c+10)/1=26/c`

Multiply both sides by c

`c^2+10c=26`

Subtract 26 from both sides

`c^2+10x-26=0 larr" A quadratic"`

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

In its normally remembered form we have:
`ax^2+bx+c=0 -> x=(-b+-sqrt(b^2-4ac))/(2a)`

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

`c=(-10+-sqrt(10^2-4(1)(-26)))/(2(1))`

`c=-5+-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+-sqrt(2^2xx51)/2`

`c=5+-(cancel(2)sqrt(51))/cancel(2)`

`c=5+-sqrt(51)`