SOLUTION: what is the square root of a^2-10a+25

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Question 127528: what is the square root of
a^2-10a+25

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
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a%5E2+-+10a+%2B+25
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You can tell this is a perfect square trinomial by dividing the multiplier of the "a" term
by 2 and then squaring it. If the answer is +25 then the given trinomial is a perfect square.
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In this problem you divide the -10 by 2 to get -5 and when you square -5 you get +25 ... just
as the problem already has. So the given trinomial is a perfect square.
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And the two factors will have half of -10 as their numerical part.
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So we can write:
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a%5E2+-+10a+%2B+25+=+%28a+-5%29%2A%28a+-+5%29+=+%28a-5%29%5E2
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Therefore, the square root of the given trinomial is the square root of:
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%28a+-+5%29%5E2
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and when you take the square root of this quantity, the answer is just:
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a+-+5
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That's the answer to your problem. Hope this helps you to see a way to get this answer.
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