Results 1 to 9 of 9

Math Help - Finding the values of p and q

  1. #1
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468

    Finding the values of p and q

    x^2-8x+23-(x-p)^2+q for all values of x

    find the value of p and q

    I did

    (x^2-8x+16)+7=...\rightarrow (x+4)^2+7=(x+p)^2+q

    I'm stuck.. what should I do now, square route the whole thing?

    edit:

    whittled it down to
    2x*2+23=p^2+q

    what now?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    The first expression you typed, is it supposed to be an equation?

    x^2-8x+23=(x-p)^2+q

    If so, you can expand on the right hand side.

    Giving x^2-8x+23=x^2 -2px + p^2 + q

    Then, equating the coefficients, you see that:

    -8 = -2p

    and

    23 = p^2+q

    Or, you can take directly from where you reached;

    (x+4)^2+7=(x+p)^2+q

    and equate the coefficients.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    Thanks, so many people have told me about the coefficient and I've looked on sites but I still don't understand how to calculate it.

    All I've learned is that comparing the coefficient is along the lines of ax^2+bx+c=a+b+c
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Okay, if you have

    ax^2 + bx + c = px^2 + qx + r

    Then, a = p because both a and p are terms in x^2
    b = q since both b and q are terms in x,
    and c = r since both are independent terms.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    That's what I was thinking about when I was doing the equation but I was unsure because it seems too perfect. Can you show me a proof?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    I'm not good at proving things

    Have you done vector coordinates yet?

    If so, you should know, for example that if you have a vector (ai + bj) and another vector (xi + yj)

    Given both are the same vectors, mean that ai + bj = xi + yj

    Automatically, a will be equal to x and b will be equal to y.

    I know, it's not a very convincing way, but I've learned it through practice
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    Quote Originally Posted by Unknown008 View Post
    I'm not good at proving things

    Have you done vector coordinates yet?

    If so, you should know, for example that if you have a vector (ai + bj) and another vector (xi + yj)

    Given both are the same vectors, mean that ai + bj = xi + yj

    Automatically, a will be equal to x and b will be equal to y.

    I know, it's not a very convincing way, but I've learned it through practice

    I hate having to merely accept things >.>

    yes I have studied them
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Ok, I thought more about it during the day. I'll use the method of completing the squares to prove it.

    Let's have a look at:

    ax^2 + bx + c= px^2 + qx + r

    I have to prove that a = p, b = q and c = r.

    (a-p)x^2 + (b-q)x + (c-r)= 0

    (a-p)[x^2 + \frac{(b-q)}{(a-p)}x + \frac{(c-r)}{(a-p)}]= 0

    Now, completing the square;

    (a-p)[(x + \frac{(b-q)}{2(a-p)}x)^2 -  (\frac{(b-q)}{2(a-p)})^2 + \frac{(c-r)}{a-p}]= 0

    So far so good.

    For the equation to be true, either (a-p) = 0, or (x + \frac{(b-q)}{2(a-p)}x)^2 -  (\frac{(b-q)}{2(a-p)})^2 + \frac{(c-r)}{a-p} = 0

    I solve the first one.

    a - p = 0

    Therefore, a = p (1st shown)

    Using that in the initial equation given,

    ax^2 + bx + c= px^2 + qx + r

    Since a = p,

    bx + c= qx + r

    I put all the terms on the same side again.

    (b - q)x + (c-r)= 0

    (b - q)[x + \frac{(c-r)}{(b-q)}]= 0

    Hence, b - q = 0, and b = q. (2nd shown)

    Again, using this new result in the previous equation,

    bx + c= qx + r

    We get:

    c = r

    I hope this one satisfies you
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    Wow that really is something. Thanks for the effort you put in! It really settles my mind :P
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the values of a, b and c.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 16th 2011, 08:35 PM
  2. Finding the values of a and b
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 18th 2011, 01:46 AM
  3. Finding max min values
    Posted in the Math Software Forum
    Replies: 6
    Last Post: December 8th 2009, 01:18 PM
  4. Finding Values of A and B
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 17th 2009, 08:40 PM
  5. Finding values of x
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 7th 2009, 08:49 AM

Search Tags


/mathhelpforum @mathhelpforum