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

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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) About Me  (Show Source):
You can put this solution on YOUR website!
To find the number of real solutions, simply use the discriminant formula D=b%5E2-4ac. If D%3E0, there are 2 real solutions. If D=0, there's only one real solutions. Finally, if D%3C0, then there are no real solutions.



From x%5E2-10x%2B25 we can see that a=1, b=-10, and c=25


D=b%5E2-4ac Start with the discriminant formula.


D=%28-10%29%5E2-4%281%29%2825%29 Plug in a=1, b=-10, and c=25


D=100-4%281%29%2825%29 Square -10 to get 100


D=100-100 Multiply 4%281%29%2825%29 to get %284%29%2825%29=100


D=0 Subtract 100 from 100 to get 0


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

Answer by stanbon(75887) About Me  (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?
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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.