Results 1 to 9 of 9

Math Help - Multiplying by conjugate

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    35

    Multiplying by conjugate

    4-sqrt(s)/(s-16)


    s=16

    i keep multiplying this by 4+sqrt(s) and keep coming up with -12/0
    the book is telling me that it's -1/8... how is this...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by NotEinstein View Post
    4-sqrt(s)/(s-16)


    s=16

    i keep multiplying this by 4+sqrt(s) and keep coming up with -12/0
    the book is telling me that it's -1/8... how is this...
    Did you factorise the numerator ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    35
    Quote Originally Posted by Moo View Post
    Did you factorise the numerator ?
    I'm all jumbled right now.. now i'm getting 0/0... *cries*
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Please show your steps and then we can help you locate your mistake.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    Posts
    35
    4-sqrt(s)/(s-16) * (4+sqrt(s)/4+sqrt(s))

    16-s/(s-16)(4+sqrt(s))

    0/ (4s+s(sqrt(s))-64-(16(sqrt(s)))

     0/ (64 + 64 - 64 - 64)

    0/0...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    I fixed your Latex code. Take a look so you know how to write fractions now

    Oh no! You changed it back to wrong.... sigh. When writing fractions in Latex, use \frac{numerator}{denominator}

    I spotted your error. You don't need to plug in s=16 and get 0 in the numerator. 16-s and s-16 are only a negative sign factored out away from equal.

    \frac{16-s}{(s-16)(4+sqrt(s))} = \frac{16-s}{-(16-s)(4+\sqrt{s})}.

    Now the 16-s cancels and you are left with -\frac{1}{4+\sqrt{s}}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    \frac{4-\sqrt{s}}{s-16} \bigg(\frac{4+\sqrt{s}}{4+\sqrt{s}}\bigg)

    \frac{16-s}{s-16}\bigg(\frac{1}{4+\sqrt{s}}\bigg)

    \frac{0}{4s+s\sqrt{s}-64-16\sqrt{s}}

    Ok I fixed your latex
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Sep 2008
    Posts
    35
    Quote Originally Posted by Jameson View Post
    I fixed your Latex code. Take a look so you know how to write fractions now

    Oh no! You changed it back to wrong.... sigh. When writing fractions in Latex, use \frac{numerator}{denominator}

    I spotted your error. You don't need to plug in s=16 and get 0 in the numerator. 16-s and s-16 are only a negative sign factored out away from equal.

    \frac{16-s}{(s-16)(4+sqrt(s))} = \frac{16-s}{-(16-s)(4+\sqrt{s})}.

    Now the 16-s cancels and you are left with -\frac{1}{4+\sqrt{s}}

    so you can just do that? just toss in a negative like that? geeezz... no algebra for 5 years really messes with my head
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    I wouldn't call it "tossing in a negative".

    Here, think about it. Let's say you have (a-b). Now if you factor out a -1 from both terms, what do you get? -1(-a+b) or -1(b-a) or -(b-a). If you distribute the -1 back through both terms you get your original sum.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. What is a conjugate?
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: July 27th 2010, 09:48 AM
  2. conjugate
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 11th 2009, 11:11 AM
  3. Conjugate?
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: September 14th 2009, 05:50 AM
  4. Solving a limit by multiplying by conjugate
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 19th 2008, 07:32 AM
  5. about conjugate
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 30th 2005, 07:12 PM

Search Tags


/mathhelpforum @mathhelpforum