SOLUTION: Here's a question from my teacher. Solve this equation by the means of your choice: (5m – 1)^2 = 25 I started out by moving the 25 to the left side---- (5m)^2-25=0 From ther

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Here's a question from my teacher. Solve this equation by the means of your choice: (5m – 1)^2 = 25 I started out by moving the 25 to the left side---- (5m)^2-25=0 From ther      Log On


   

Question 133464: Here's a question from my teacher.
Solve this equation by the means of your choice: (5m – 1)^2 = 25
I started out by moving the 25 to the left side---- (5m)^2-25=0
From there I get 25m-24=0, but I get stuck after that????
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
You started by making a small error. Moving the constant term to the left side of the equation should have resulted in %285m-1%29%5E2-25=0, then you would have to expand the binomial to get:

25m%5E2-10m%2B1-25=0
25m%5E2-10m%2B24=0

And this factors to:
%285m-6%29%285m%2B4%29=0, so

m=6%2F5 or m=%28-4%29%2F5.

Whew! That was a lot of work. But there is an easier way:

%285m+-+1%29%5E2+=+25

Just take the square root of both sides:
5m-1=5 or 5m-1=-5

m=6%2F5 or m=%28-4%29%2F5