Results 1 to 8 of 8

Math Help - Help again with Modulus & Irrational Equations

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    12

    Help again with Modulus & Irrational Equations

    Can anyone help with this as I am stumped:

    1 + square root of x = square root of 3(x-1)

    The answer is x=4 but I can't get it.
    Also can you tell me how to do the square root symbol etc when posting a thread.

    Thanks and Kind Regards
    Larsson
    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
    Hello,

    Square the whole thing...

    Then simplify what you can and subsitute u=\sqrt{x}




    For the square root : between [math ] and [/tex] do \sqrt
    See the code by clicking on my square root
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    12
    Thanks for your help moo, unfortunately I am totally lost with regard to the equation but thank you for helping me with the symbols!!!!

    Regards
    Larsson
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by larsson View Post
    Can anyone help with this as I am stumped:

    1 + square root of x = square root of 3(x-1)

    The answer is x=4 but I can't get it.
    Also can you tell me how to do the square root symbol etc when posting a thread.

    Thanks and Kind Regards
    Larsson
    as Moo directed.

    1 + \sqrt{x} = \sqrt{3(x - 1)} ..........square both sides

    \Rightarrow (1 + \sqrt{x})^2 = (\sqrt{3(x - 1)})^2 .........simplify

    \Rightarrow 1 + 2 \sqrt{x} + x = 3(x - 1)

    can you continue?\
    Follow Math Help Forum on Facebook and Google+

  5. #5
    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
    Ok, i'll go through it

    1+\sqrt{x}=\sqrt{3(x-1)}

    If we square both sides :

    \underbrace{(1+\sqrt{x})^2}_{1+x+2\sqrt{x}}=3(x-1)=3x-3

    1+x+2 \sqrt{x}=3x-3

    2 \sqrt{x}=2x-4

    x-\sqrt{x}-2=0

    Substitute u=\sqrt{x}

    x=(\sqrt{x})^2=u^2

    So it's the same as solving for u (positive) in :

    u^2-u-2=0

    This is (u-2)(u+1)=0

    So u is either equal to 2 or to -1.

    But u=\sqrt{x}>0

    So u=2.

    Thus x=...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2008
    Posts
    12


    If we square both sides :






    Its up to here that I could get to and understand. The rest of it has me totally lost, I am still not sure how the answer is 4.

    I really appreciate all the help you have all given me and I hope I am not annoying you all

    Regards
    Larsson
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by larsson View Post


    If we square both sides :






    Its up to here that I could get to and understand. The rest of it has me totally lost, I am still not sure how the answer is 4.

    I really appreciate all the help you have all given me and I hope I am not annoying you all

    Regards
    Larsson
    Moo got the next line by bring everything to one side and dividing through by 2

    from there, she then made the said substitution, and the rest is algebra. at the end, she solved for u. But we know that x = u^2, so ...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2008
    Posts
    12
    Thanks a lot Jhevon and Moo for all your help I think I understand it now. I am currently trying to teach myself Maths and it was the highlighted bit below that I never had seen before


    But we know that x = u^2,

    Thanks and Regards
    Larsson
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simultaneous equations with a modulus
    Posted in the Algebra Forum
    Replies: 7
    Last Post: May 5th 2011, 01:01 PM
  2. Irrational equations
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: November 24th 2010, 08:41 AM
  3. Irrational Equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 26th 2009, 02:46 AM
  4. Irrational Equations
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: October 5th 2008, 03:41 AM
  5. Modulus and Irrational Equations
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 15th 2008, 04:42 AM

Search Tags


/mathhelpforum @mathhelpforum