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Math Help - Find the exact value... radicals.

  1. #1
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    Find the exact value... radicals.

    Having a hard time finding the answer to the problem. Can someone help me with this. (6-1)^2 + (15-3)^2 all of this is under a radical sign. I think the answer is 16.46. But I am not sure. I started out by squaring the two sets getting 36+1 +225+9 which equals 271 under a radical sign giving me the answer 16.46. Is this correct
    Last edited by modecasec; April 23rd 2012 at 08:25 AM.
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  2. #2
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    Re: Find the exact value... radicals.

    Quote Originally Posted by modecasec View Post
    Having a hard time finding the answer to the problem. Can someone help me with this. (6-1) + 15-3)2 all of this is under a ridical sign. I think the answer is 16.46. But I am not sure. I started out by squaring the two sets getting 36+1 +225+9 which equals 271 under a radical sign giving me the answer 16.46. Is this correct
    Hi modecasec,

    There's a problem with your notation. Your parentheses aren't balanced. You're missing one left parenthesis. Check you problem and edit your post.

    What you have is \sqrt{(6-1)+15-3)^2 and that is incorrect.

    It could be \sqrt{((6-1)+(15-3))^2}=17, or

    \sqrt{(6-1)+(15-3)^2}=\sqrt{149}. I'm just guessing at what you intended, so edit your post.


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  3. #3
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    Re: Find the exact value... radicals.

    yes I messed that up. The two sets in parenthesis are raised to the secodn degree. what do you get now.
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  4. #4
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    Re: Find the exact value... radicals.

    Hello, modecasec!

    It's hard to read what you typed, but I'll take a guess . . .


    \sqrt{(6-1)^2 + (15-3)^2}

    First of all, be careful . . . Don't make up your own rules!
    . . (6-1)^2 is not equal to 6^2+1^2

    Look at what we have:
    . . (6-1)^2 \:=\:5^2 \:=\:25
    . . (15-3)^2 \:=\:12^2 \:=\: 144


    So we have: . \sqrt{25 + 144} \:=\:\sqrt{169} \:=\:\boxed{13} . **


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    Another opportunity to make a fatal error . . .

    . . \sqrt{25 + 144} .is not equal to . \sqrt{25} + \sqrt{144}


    Very tempting, isn't it? .Both 25 and 144 are squares.
    Well, of course they are!
    . . They just came from 5^2 and 12^2, didn't they?

    This tempting error appears every time we use the Pythagorean Theorem:
    . . a^2 + b^2 \:=\:c^2
    Think about it.

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