Find all values of #c# such that #c/(c-5) = 4/(c-4)#. Help, Please?

1 Answer

There are no real values of #c# that will make this equation work. However, there are imaginary solutions: #c=-4+2i, -4-2i#

Explanation:

#c/(c-5)=4/(c-4)#

Cross multiply:

#c(c-4)=4(c-5)#

Distribute:

#c^2-4c=4c-20#

Set equal to 0:

#c^2-8c+20=0#

I'll use the quadratic formula:

# x = (-b \pm sqrt(b^2-4ac)) / (2a) #

with #a=1, b=-8,c=20#

# x = (-8 \pm sqrt((-8)^2-4(1)(20))) / (2(1)) #

# x = (-8 \pm sqrt(64-80)) / 2 #

There are no real solutions to this problem. There are, however, imaginary solutions:

# x = (-8 \pm sqrt(-16)) / 2 =(-8pm4i)/2#

#x=-4+2i, -4-2i#