SOLUTION: how do you find the number of real solutions for this equation: x^2-10x+25=0?

Question 171875: how do you find the number of real solutions for this equation: x^2-10x+25=0?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
To find the number of real solutions, simply use the discriminant formula . If , there are 2 real solutions. If , there's only one real solutions. Finally, if , then there are no real solutions.

From we can see that , , and

Start with the discriminant formula.

Plug in , , and

Square to get

Multiply to get

Subtract from to get

Since the discriminant is equal to zero, this means that there is one real solution.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
how do you find the number of real solutions for this equation: x^2-10x+25=0?
--------
You evaluate the discriminant: b^2-4ac
Your discriminant = 10^2 - 4*1*25 = 100-100 = 0
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So the equation has two equal Real Number solutions.
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what are they?
Factor your problem and ou get:
(x-5)^2 = 0
So x = 5 with multiplicity two.
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Cheers,
Stan H.
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