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Math Help - Equation needs URGENT checking

  1. #1
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    Equation needs URGENT checking

    Hello, i was preparing for a test i have tomorrow so was doing review questions in my text book. I came across this one question where i knew what to do but got the wrong answer, must be a mistake in my working out
    Question: the distance between (2,-7) and (a,-2) is \sqrt{41}units. Find the value of a if it is positive.

    Edit: New to using Latex - sorry for any inconvenience
    AND
    The answer i have been obtaining is a=4

    The formula to use was d = \sqrt(x2-x1)^2+(y2-y1)^2
    in this case d being \sqrt 41
    the actual answer is a= 6..

     \sqrt41 = \sqrt{(a-2)^2+(-2+7)^2}<br />
\Rightarrow \sqrt41 = \sqrt{(a^2-4a+29)}<br />
\Rightarrow 41 = a^2-4a+29<br />
     \Rightarrow 12 = a^2-4a<br />
\Rightarrow 12/a = a-4<br />
    <br />
\Rightarrow 12/a+4 = a<br />
\Rightarrow 12+4 = a^2<br />
    <br />
\Rightarrow 16= a^2<br />
\Rightarrow a=4<br />
    Last edited by 99.95; February 21st 2010 at 03:51 AM.
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  2. #2
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    Quote Originally Posted by 99.95 View Post
    Hello, i was preparing for a test i have tomorrow so was doing review questions in my text book. I came across this one question where i knew what to do but got the wrong answer, must be a mistake in my working out
    Question: the distance between (2,-7) and (a,-2) is [tex]\sqrt{41}[\math]units. Find the value of a if it is positive.

    Edit: New to using Latex, why wont it show upo
    because you end with /math, not \math.

    AND
    The answer i have been obtaining is a=4

    The formula to use was d = \sqrt(x2-x1)^2+(y2-y1)^2
    in this case d being \sqrt 41
    the actual answer is a= 6..
    What do you want? If you know the answer is 6, then you know your answer is wrong. If you want us to check your work and see where you went wrong, you will need to show your work!
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    Quote Originally Posted by HallsofIvy View Post
    because you end with /math, not \math.


    What do you want? If you know the answer is 6, then you know your answer is wrong. If you want us to check your work and see where you went wrong, you will need to show your work!
    ok ill post it now, yea i want you to check and see where i went wrong

     \sqrt41 = \sqrt{(a-2)^2+(-2+7)^2}<br />
\Rightarrow \sqrt41 = \sqrt{(a^2-4a+29)}<br />
\Rightarrow 41 = a^2-4a+29<br />
\Rightarrow 12 = a^2-4a<br />
\Rightarrow 12/a = a-4<br />
\Rightarrow 12/a+4 = a<br />
\Rightarrow 12+4 = a^2<br />
\Rightarrow 16= a^2<br />
\Rightarrow a=4<br />
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  4. #4
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     \sqrt41 = \sqrt{(a-2)^2+(-2+7)^2}<br />
\Rightarrow \sqrt41 = \sqrt{(a^2-4a+29)}<br />
\Rightarrow 41 = a^2-4a+29<br />
     \Rightarrow 12 = a^2-4a<br />
\Rightarrow 12/a = a-4<br />
    <br />
\Rightarrow 12/a+4 = a<br />
\Rightarrow 12+4 = a^2<br />
    <br />
\Rightarrow 16= a^2<br />
\Rightarrow a=4<br />
    Last edited by 99.95; February 21st 2010 at 03:51 AM.
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  5. #5
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    WHAT are you doing?!
    You have a^2 - 4a + 29 = 41; so:
    a^2 - 4a - 12 = 0
    (a - 6)(a + 2) = 0
    It's that simple
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  6. #6
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    Quote Originally Posted by Wilmer View Post
    WHAT are you doing?!
    You have a^2 - 4a + 29 = 41; so:
    a^2 - 4a - 12 = 0
    (a - 6)(a + 2) = 0
    It's that simple
    Wow, even though it was so simple, i would have never gotten that in a million years..
    Too rusty for my own good.
    thanks though!
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  7. #7
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    Never too late; pick a couple sites here...become an expert!

    introduction quadratics - Google Search=
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