Results 1 to 11 of 11

Math Help - Assistance required with what seems a basic simplification.

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    47

    Assistance required with what seems a basic simplification.

    My expression is -3=[(-x*pi)/(sqrt(1-x^2))]

    An online equation solver gave a solution of x=3/[sqrt(9+pi^2)] or x=.690621 but does not show the working.

    I have spent hours on this but am positively stuck!

    Can anyone help?


    Regards.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2009
    Posts
    130
    Hello!
    Use LaTeX in future for writing equations.

    -3=\frac{-\pi x}{\sqrt{1-x^2}}

    Hint:

    (-3)^2=(\frac{-\pi x}{\sqrt{1-x^2}})^2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    \displaystyle -3 = \frac{-x\times \pi}{\sqrt{1-x^2}}

    \displaystyle -3 \sqrt{1-x^2}= -x\times \pi

    \displaystyle 3 \sqrt{1-x^2}= x\times \pi

    \displaystyle 3^2 (1-x^2)= x^2\times \pi^2

    \displaystyle 9-9x^2= x^2\times \pi^2

    \displaystyle 9-9x^2- \pi^2x^2=0

    \displaystyle 9-(9+ \pi^2)x^2 =0

    \displaystyle (9+ \pi^2)x^2 =9

    \displaystyle x^2 =\frac{9}{9+ \pi^2}

    \displaystyle x =\sqrt{\frac{9}{9+ \pi^2}}
    Last edited by pickslides; January 3rd 2011 at 01:27 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2010
    Posts
    47
    Thank you very much for your replies - that one was seriously haunting me!

    Regards.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2010
    Posts
    47
    I am curious to know - what rule tells me I cannot touch the terms under the square root?

    To me sqrt(9+x^2) is 3+x but this is clearly not the case in this instance.

    Some further clarification would be much appreciated.

    Regards.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    \displaystyle \sqrt{9+x^2} \neq 3+x

    I'm not sure I follow what you are saying.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2010
    Posts
    47
    Obviously I am a bit rusty with my laws of powers etc. I thought sqrt(9+x^2) = sqrt(9) + sqrt(x^2) = 3 +x.

    As I said - a bit rusty. Further reading necessary.

    Regards.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    You may be confused with

    \displaystyle \sqrt{a\times b } = \sqrt{a}\sqrt{b}

    \displaystyle \sqrt{a+ b } \neq \sqrt{a}+\sqrt{b}
    Last edited by pickslides; January 3rd 2011 at 02:49 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2010
    Posts
    47
    In your second line did you mean sqrt{a+b }?

    I apologise - not only am I rusty at powers but am new to Latex.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jan 2011
    From
    NSW, Australia
    Posts
    15
    Generally speaking, you can only simplify a radical expression that is in factored form.

    \sqrt{9x^2} for example CAN be simplified to 3x

    The expression \sqrt{x^2 + 9} on the other hand cannot be simplfied as-is.

    This is especially apparent when you recall that (a + b)^2 \neq a^2 + b^2, because the square root would be the same as (a + b)^{\frac{1}{2}}, which by the same token, is not equal to a^{\frac{1}{2}} + b^{\frac{1}{2}}.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Mar 2010
    Posts
    47
    Great information. Thank you.

    Makes me think of BOMDAS - brackets first, then powers.

    Regards.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability assistance required
    Posted in the Statistics Forum
    Replies: 5
    Last Post: January 31st 2011, 11:20 PM
  2. Basic Truth Table help required
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: December 10th 2010, 06:41 PM
  3. Replies: 6
    Last Post: December 8th 2010, 02:18 PM
  4. Basic equation assistance...
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 8th 2009, 07:40 AM
  5. Very basic limit help required!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 11th 2009, 06:59 AM

Search Tags


/mathhelpforum @mathhelpforum