Results 1 to 5 of 5
Like Tree2Thanks
  • 1 Post By agentmulder
  • 1 Post By ibdutt

Math Help - Cannot for the life of me follow this simplification

  1. #1
    Junior Member
    Joined
    Mar 2013
    From
    New Zealand
    Posts
    35

    Cannot for the life of me follow this simplification



    I understand how the sqrt comes down, from the right half of the top, but how does the top left part lose the square root?

    After simplifying, I can get

    sqrt(x^2+1) - x^2 / (x^2+1)^(3/2)

    However, the answer that all online calcs give and my tutorial notes give is

    1/ (x^2+1)^(3/2)

    And I can see where that is going from the working up the top, but can't figure out how the sqrt(x^2+1) turns into (x^2+1)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member agentmulder's Avatar
    Joined
    Aug 2011
    From
    42nd parallel, North America
    Posts
    105
    Thanks
    33

    Re: Cannot for the life of me follow this simplification

    One way to see this is to factor, first cancel the 2 with 1/2 at top right then factor

     \frac{(x^2 + 1)^{-\frac{1}{2}}[(x^2 + 1) - x^2]}{x^2 + 1}

    Notice if you distribute, the bases are the same so you add exponents and keep the base , -1/2 + 1 = 1/2 so we're good

    Next simplify within brackets = 1 then bring what's left to the denominator so it's exponent is positive, finally keep the same base and add the exponents. Let me know if you need more details.

    Last edited by agentmulder; May 13th 2013 at 01:19 AM.
    Thanks from lukasaurus
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2013
    From
    New Zealand
    Posts
    35

    Re: Cannot for the life of me follow this simplification

    Yep,I definitely need more details with that first factoring. I have been sitting, trying to work this out for over two hours
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jul 2012
    From
    INDIA
    Posts
    829
    Thanks
    209

    Re: Cannot for the life of me follow this simplification

    Cannot for the life of me follow this simplification-13-may-2.png
    Thanks from lukasaurus
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member agentmulder's Avatar
    Joined
    Aug 2011
    From
    42nd parallel, North America
    Posts
    105
    Thanks
    33

    Re: Cannot for the life of me follow this simplification

    Quote Originally Posted by lukasaurus View Post
    Yep,I definitely need more details with that first factoring. I have been sitting, trying to work this out for over two hours
    Let's just look at only the numerator for a moment , after canceling 1/2 with 2 and let's represent the radical with an equivalent exponent.

     (x^2 + 1)^{\frac{1}{2}} - x^2(x^2 + 1)^{-\frac{1}{2}}

    is it easier to see the same base now?

    We must factor from both terms because we have subtraction between the 2 terms.

    Factoring from the term on the right is easy, just pull it out, factoring from the term on the left is a bit harder but if you can understand it you save time and space. Essentialy , to figure out what exponent to put you subtract the exponents

    We identified the common base

     (x^2 + 1)

    We decide to factor out

     (x^2 + 1)^{-\frac{1}{2}}

    So far we have

     (x^2 + 1)^{-\frac{1}{2}}[ \ \ \ \ \ \          - x^2]

    what do we put in the space? Well, it's going to be the same base with an exponent of...

     \frac{1}{2} - (- \frac{1}{2} ) = 1

    the positive 1/2 is the exponent on the base that is getting factored the -1/2 is the exponent on the base that is doing the factoring.

     (x^2 + 1)^{-\frac{1}{2}}[(x^2 + 1)^1 - x^2]

    or simply

     (x^2 + 1)^{-\frac{1}{2}}[x^2 + 1 - x^2]

    The rest of it is obvious, i hope to have helped you understand the factoring. Let me know if anything is not clear.

    Last edited by agentmulder; May 13th 2013 at 02:37 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 1st 2012, 01:48 PM
  2. Do we follow BEDMAS or ...?
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 11th 2012, 05:08 PM
  3. Real life problem My life plz help
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: September 30th 2010, 02:44 AM
  4. Does this necessarily follow?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 22nd 2010, 02:33 PM
  5. Follow the bouncing limit
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 8th 2009, 09:12 PM

Search Tags


/mathhelpforum @mathhelpforum