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Math Help - Find integers

  1. #1
    Member GAdams's Avatar
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    Find integers

    What do they mean by 'for all x' in (a)?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by GAdams View Post
    What do they mean by 'for all x' in (a)?
    It means find p and q such that whatever value x takes:

    <br />
x^2-8x=(x-p)^2+q<br />

    That is find p and q so that this is an identity.

    To do this expand the right hand side, and equate the coefficients of like
    powers of x on both sides of the equation.

    RonL
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  3. #3
    MHF Contributor red_dog's Avatar
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    a) x^2-8x=(x-p)^2+q\Leftrightarrow x^2-8x=x^2-2px+p^2+q,\forall x\in\mathbf{R}. That means all coefficients of both members are equal.
    Then \left\{\begin{array}{c}-2p=-8\\p^2+q=0\end{array}\right.\Rightarrow p=4,q=-16.
    b) The equation can be written
    x^2-8x+16+y^2=16\Leftrightarrow (x-4)^2+y^2=16.
    c) The graph is a circle with center C(4,0) and radius r=4.
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  4. #4
    Member GAdams's Avatar
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    Quote Originally Posted by red_dog View Post
    a) x^2-8x=(x-p)^2+q\Leftrightarrow x^2-8x=x^2-2px+p^2+q,\forall x\in\mathbf{R}. That means all coefficients of both members are equal.
    Then \left\{\begin{array}{c}-2p=-8\\p^2+q=0\end{array}\right.\Rightarrow p=4,q=-16.
    b) The equation can be written
    x^2-8x+16+y^2=16\Leftrightarrow (x-4)^2+y^2=16.
    c) The graph is a circle with center C(4,0) and radius r=4.
    What does the upside down A and the R mean?
    I don't understand how you got part 'b'.
    Thanks.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    What does the upside down A and the R mean?
    I don't understand how you got part 'b'.
    Thanks.
    the upside down A means "for all" so he's saying "for all x"

    the "R" means the set of real numbers, so he is really saying "for all real x" or "for all x in the set of real numbers"
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  6. #6
    Member GAdams's Avatar
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    Thanks, but how can you write the equation as it's written in part (b)?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    Thanks, but how can you write the equation as it's written in part (b)?
    complete the square for the x-term, you know how to complete the square right?
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  8. #8
    Member GAdams's Avatar
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    Ok, I got part (b) thus:

    x^2 + y^2 -8x =0

    by completing the square (which I just learnt how to do)

    x^2 + y^2 -8x + 16 = 16

    (x-4)^2 + y^2 = 16

    Part (c) is clear


    Part (a) is still confusing me. I don't understand it all.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    Part (a) is still confusing me. I don't understand it all.
    I think red_dog did an excellent job on this question. However, i shall take baby steps and leave out the formalism, hoefully you can get the concept from this.


    We have x^2 - 8x = (x - p)^2 + q

    we want this to be true no matter what x we choose. we will do the classic move of equating coefficients to accomplish this. Let's expand the right side, we get:

    x^2 - 8x = x^2 - 2xp + p^2 + q

    Now group like powers of x

    \Rightarrow x^2 - 8x = x^2 - (2p)x + \left( p^2 + q \right)

    so now we match up the coefficients. since the two sides are equal, we must have the same number of each power of x on both sides.

    we have 1 x^2 on each side, so we're good there

    on the left we have -8x on the right we have -2px, they must be equal, so we must have:

    -8 = -2p \implies \boxed { p = 4 }

    on the left we have 0 as the constant term, on the right we have p^2 + q as the constant term. since we want the constants to be the same as well for the sake of equality, we must have:

    0 = p^2 + q

    \Rightarrow 0 = 16 + q ..........since p = 4

    \Rightarrow \boxed { q = -16 }

    so those are the two values we need. so no matter what x is, those values for p and q will work


    did you get it?
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  10. #10
    Member GAdams's Avatar
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    Got it!! Thanks. Great explanation!
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