<PRE><B>Question 171875</B>
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.</PRE>