Results 1 to 10 of 10

Math Help - quadratic equation

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    Manchester
    Posts
    15

    quadratic equation

    have i done this correctly? was unsure about the 18/2 part?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1

    Re: quadratic equation

    Given an equation of the form

    ax^2 + bx +c=0,

    we can find solutions for x using the formula

    x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

    Check that you have used this formula correctly. An easy way to tell if you have the right answer is to substitute your values for x back into the original equation to see if you get the right answer i.e try and work out whether (4.45)^2+2(4.45)-3.5 really equals zero. If not then you need to check your calculations carefully.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,168
    Thanks
    71

    Re: quadratic equation

    You have to be more careful; you should get:

    x = [-2 +- SQRT(18)] / 2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2012
    From
    Manchester
    Posts
    15

    Re: quadratic equation

    I always seem to make careless mistakes, so it should have been 2.12 rather than 2.45 !!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: quadratic equation

    there are two problems with your solution:

    1) EVERYTHING goes over the 2, not just "the square root part"

    2)you should wind up with:

    \frac{-2 \pm \sqrt{18}}{2}

    it's ok to leave it like this, but if you want to simplify:

    \sqrt{18} = \sqrt{2}\sqrt{9} = 3\sqrt{2}.

    neither:

    \frac{-2 + 3\sqrt{2}}{2} nor \frac{-2 - 3\sqrt{2}}{2} are even close to either one of your "proposed" answers 2.12 or 2.45. there is no √6 involved in the correct solution.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2012
    From
    Manchester
    Posts
    15

    Re: quadratic equation

    Yeh that's what I mean about making careless mistakes, so really it would be 0.12 and -4.12 ? Thanks!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,168
    Thanks
    71

    Re: quadratic equation

    NO! Try again ..... CAREFULLY!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Aug 2012
    From
    Manchester
    Posts
    15

    Re: quadratic equation

    1.12 and -3.12?

    Eeek, hope this is right now :/
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,168
    Thanks
    71

    Re: quadratic equation

    YES.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Aug 2012
    From
    Manchester
    Posts
    15

    Re: quadratic equation

    Woooo haha thanks Wilmer!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic Equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 17th 2010, 12:23 AM
  2. Replies: 3
    Last Post: April 25th 2010, 04:53 PM
  3. Quadratic Equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 24th 2009, 04:04 AM
  4. quadratic equation?
    Posted in the Algebra Forum
    Replies: 7
    Last Post: November 18th 2009, 03:39 AM
  5. One more quadratic equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 15th 2009, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum