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

2 Answers
May 18, 2018

Answer:

#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)#

May 18, 2018

Answer:

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.

Tony B

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

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

#c=5+-sqrt(51)#