SOLUTION: Find b^2 - 4ac and the number of real solutions: 4m^2 + 25 = 20m The book shows an answer of (0, 1) I have tried to work it by changing to a quadratic and using the discri

Question 147347This question is from textbook Elementary and Intermediate Algebra
: Find b^2 - 4ac and the number of real solutions:
4m^2 + 25 = 20m
The book shows an answer of (0, 1)
I have tried to work it by changing to a quadratic and using the discriminant to figure the answer, but I end up with m = 5/2
I would greatly appreciate if someone can tell me how to work this problem. I am completely stumped. Thanks! This question is from textbook Elementary and Intermediate Algebra

Answer by nabla(475) (Show Source): You can put this solution on YOUR website!
Rewrite as
4m^2-20m+25=0
where
am^2+bm+c=0 is the form we're interested in.
so
b^2-4ac=(-20)^2-4*4*25=400-400=0

Since the discriminant is 0 we shall have 1 solution. You can see why this is so from the quadratic formula:

If we eliminate the root due to a zero, the +/- is irrelevant and leaves only -b/2a as the solution of x.

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