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Math Help - Algebra.

  1. #1
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    Algebra.

    For all the values of x,

    x^2 - 6x + 15 = (x-p)^2 + q

    Find the values of p and q

    p=

    q=

    Please. Thankyou
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  2. #2
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    Quote Originally Posted by alexhill View Post
    For all the values of x,

    x^2 - 6x + 15 = (x-p)^2 + q

    Find the values of p and q

    p=

    q=

    Please. Thankyou
    What you need to do is expand the right side first. (x-p)^2 + q = x^2 -2px + p^2 + q.

    Equating coefficients gives you -2p = -6 and p^2 + q = 15. You should be able to solve it from there.
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  3. #3
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    For all the values of x,

    x^2 - 6x + 15 = (x-p)^2 + q

    Find the values of p and q
    Oki doki. This is actually a maths thing called "Completing the Square" - as you'll probably find out, it's a nice way of solving some types of maths problems. Anyway, you really want to be making the left hand side look like the right hand side so you can see what the numbers are.

    x^2 - 6x + 15 = (x-p)^2 + q

    First of all, you have that (x-p)(x-p) on the right which needs to be multiplied out. So let's do that (I hope you know how to multiply out those brackets - if not, then say so and I'm sure someone will help)

    x^2 - 6x + 15 = x^2 - 2px +p^2 + q

    Take away x^2 from both sides

    -6x + 15 = -2p x + p^2 +q

    Now, how many xs are there on the left? -6. How many are there on the right? -2p.

    How many plain numbers with no xs are there on the left? 15. How many on the right? p^2 + q.

    Since the left hand side EQUALS the right hand side, we can say:

    -2p = -6

    and

    15 = p^2 + q

    Read that again if it didn't make sense - that's a method called "equating coefficients" and it's something that comes in useful a lot.

    Good luck.

    - - - - - -
    Edit:

    Beaten to it by icemanfan, sorry.
    Last edited by Fedex; April 25th 2008 at 09:58 AM. Reason: beaten to it
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  4. #4
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    .

    Yeah, that makes sense. Didn't understand the equating coefficients thing. I'll learn that, thanks.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    alternatively, you could complete the square on the left, so you would know p and q right away
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alexhill View Post
    Didn't understand the equating coefficients thing. I'll learn that, thanks.
    the main idea with this is that you want equality. that means you have to have the same number of things on both sides of the equation. if you had 5x's on one side of an equation and 2x's on the other, they wouldn't be equal of course. equating coefficients simply means, you want to have the same number of x's, x^2's etc on either side of the equation.

    i think Fedex did a nice job of explaining further. re-read his post to see if it is clearer now
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