Results 1 to 3 of 3

Math Help - Why are you my answers wrong?

  1. #1
    Member
    Joined
    May 2008
    Posts
    112

    Why are my answers wrong?

    1) A quadratic equation in two variables always have two possible solutions in any number system.

    My answer: False, it has no solution when a = 0.

    2) All equations in two variables, such that both variables have maximum exponent as 2 and minimum as zero represents an equation of a circle.

    My answer: False, in equation of a circle, when exponent is 0, (x-h)^0+(y-k)^0=r^2 is not a circle


    3) Find the composition of the function as indicated, find domain of composite function and explain your answer(s).

    f(x) = ( (4x^2) - 2) / (-4 + 8x)
    g(x) = ( (9x^2) -9) / (-12+6x)
    Find fog and fof
    Last edited by mwok; October 13th 2008 at 08:39 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Aug 2008
    Posts
    530
    Quote Originally Posted by mwok View Post
    1)

    3) Find the composition of the function as indicated, find domain of composite function and explain your answer(s).

    f(x) = ( (4x^2) - 2) / (-4 + 8x)
    g(x) = ( (9x^2) -9) / (-12+6x)
    Find fog and fof
    f\left( x \right) = \frac{{4x^2  - 2}}<br />
{{ - 4 + 8x}} = \frac{{2\left( {2x^2  - 1} \right)}}<br />
{{2\left( {4x - 2} \right)}} = \frac{{2x^2  - 1}}<br />
{{4x - 2}} \hfill \\

    g\left( x \right) = \frac{{9x^2  - 9}}<br />
{{ - 12 + 6x}} = \frac{{3x^2  - 3}}<br />
{{2x - 4}} \hfill \\

    fog = f\left( {g\left( x \right)} \right) = \frac{{2\left( {\frac{{3x^2  - 3}}<br />
{{2x - 4}}} \right)^2  - 1}}<br />
{{4\left( {\frac{{3x^2  - 3}}<br />
{{2x - 4}}} \right) - 2}} \hfill \\

    fog = \frac{{2\left( {3x^2  - 3} \right)^2  - \left( {2x - 4} \right)^2 }}<br />
{{4\left( {3x^2  - 3} \right)\left( {2x - 4} \right) - 2\left( {2x - 4} \right)^2 }} \hfill \\

    = \frac{{2\left( {9x^4  - 18x^2  + 9} \right) - \left( {4x^2  - 16x + 16} \right)}}<br />
{{2\left( {2x - 4} \right)\left[ {2\left( {3x^2  - 3} \right) - \left( {2x - 4} \right)} \right]}} \hfill \\

       = \frac{{18x^4  - 36x^2  + 18 - 4x^2  + 16x - 16}}<br />
{{2\left( {2x - 4} \right)\left( {6x^2  - 6 - 2x + 4} \right)}} \hfill \\

       = \frac{{18x^4  - 40x^2  + 16x + 2}}<br />
{{2\left( {2x - 4} \right)\left( {6x^2  - 2x - 2} \right)}} \hfill \\

       = \frac{{9x^4  - 20x^2  + 8x + 1}}<br />
{{\left( {2x - 4} \right)\left( {6x^2  - 2x - 2} \right)}} \hfill \\

       = \frac{{9x^4  - 20x^2  + 8x + 1}}<br />
{{4\left( {x - 2} \right)\left( {3x^2  - x - 1} \right)}} \hfill \\

      {\text{For domain, }}x \ne 2,{\text{  and  }}3x^2  - x - 1 \ne 0 \hfill \\

       \Rightarrow x \ne 2,{\text{  and  }}x \ne \frac{{1 \pm \sqrt {1^2  - 4\left( 3 \right)\left( { - 1} \right)} }}<br />
{{2\left( 3 \right)}} \hfill \\

       \Rightarrow x \ne 2,{\text{  and  }}x \ne \frac{{1 \pm \sqrt {13} }}<br />
{6} \hfill \\

    {\text{Now, }} \hfill \\

    fof = f\left( {f\left( x \right)} \right) = \frac{{2\left( {\frac{{2x^2  - 1}}<br />
{{4x - 2}}} \right)^2  - 1}}<br />
{{4\left( {\frac{{2x^2  - 1}}<br />
{{4x - 2}}} \right) - 2}} \hfill \\

    {\text{simplify further}}{\text{.}} \hfill \\
    Last edited by Shyam; October 13th 2008 at 09:54 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    (1) If a = 0, you would no longer have a quadratic which is what the question is asking for.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Are my book answers wrong?
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 19th 2010, 03:30 PM
  2. Fire Hose (Linear... Wrong Answers)
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 20th 2010, 08:46 AM
  3. Replies: 6
    Last Post: August 12th 2009, 02:15 AM
  4. Wrong answers on test, need help
    Posted in the Trigonometry Forum
    Replies: 10
    Last Post: July 22nd 2009, 09:14 PM
  5. Why are my answers wrong?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 13th 2008, 10:02 AM

Search Tags


/mathhelpforum @mathhelpforum