Results 1 to 5 of 5

Math Help - quadratic equation

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    31

    quadratic equation

    Solve by factorizing

    6x^2 -x -12=0

    Apparently it equals to zero, but im not sure how they got it, so could someone help please????
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by fvaras89 View Post
    Solve by factorizing

    6x^2 -x -12=0

    Apparently it equals to zero, but im not sure how they got it, so could someone help please????
    Do you know how to factorise 6x^2 - x - 12? See Wolfram|Alpha
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    From
    A forrest
    Posts
    162
    Awards
    1
    Quote Originally Posted by fvaras89 View Post
    Solve by factorizing

    6x^2 -x -12=0

    Apparently it equals to zero, but im not sure how they got it, so could someone help please????
    Equals to zero. That's what happens when all terms are one side to the extent of my knowledge since this one definitely does factor.

    (3x+4) (2x-3)

    6x^2 -9x+ 8x -12.
    6x^2 - x - 12. Checks out.

    You just have to realize what'll work and what won't.
    Try plugging in numbers that you think will get you 6x^2 but at the same time think of two numbers that will get you the other two numbers. Look at the factors of 12. 1, 2, 3, 4, 6, 12. Try them. 1 x 12, 2 x 6, 3 x 4 are the pairs.There are two ways to get six. 1, 2, 3, 6. 2 x 3 or 1 x 6. And it should become pretty intuitive to you once you get how factoring works.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448
    fvaras89: "Solve by factorizing, 6x^2 -x -12=0. Apparently it equals to zero, but im not sure how they got it, so could someone help please?"

    6x^2 -x -12 = (2x-3) (3x+4), but why?

    without the aid of the computer or calculator, you are left to solve by hand,

    6x^2 -x -12 = 0, can be factored in this form, (ax + b)(cx + d)

    factor of 6: 6 & 1, 3 & 2, 2 & 3.

    factor of -12: -12 & 1, -1 & 12, -6 & 2, -2 & 6, -4 & 3, -3 & 4

    Now analyze what pairing will generate 6x^2 -x -12.

    We choose 6: 2 & 3,

    and for -12: -3 & 4.

    Now we have,

    A) (2x + 4)(3x - 3) or

    B) (2x - 3)(3x + 4).

    Just check the MIDDLE term for

    A) middle term = 2x(-3) + 4(3x) = -6x + 12x = 6x, wrong result, next

    B) middle term = 2x(4) + (-3)(3x) = 8x - 9x = -x, YES!

    Now, our factorization is, 6x^2 -x -12 = (2x-3) (3x+4).

    ok - - - - -
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by pacman View Post
    fvaras89: "Solve by factorizing, 6x^2 -x -12=0. Apparently it equals to zero, but im not sure how they got it, so could someone help please?"

    6x^2 -x -12 = (2x-3) (3x+4), but why?

    without the aid of the computer or calculator, you are left to solve by hand,

    6x^2 -x -12 = 0, can be factored in this form, (ax + b)(cx + d)

    factor of 6: 6 & 1, 3 & 2, 2 & 3.

    factor of -12: -12 & 1, -1 & 12, -6 & 2, -2 & 6, -4 & 3, -3 & 4

    Now analyze what pairing will generate 6x^2 -x -12.

    We choose 6: 2 & 3,
    You have provided no explanation of what you are trying to do or why. Maths is not some magical performance where one waves ones hands does this does that and the answer magically appears.

    So in this case you explain that the rational roots theorems tells you that if this quadratic has any rational roots they are or the form of the ratio of a factor of the constant term to a factor of the coefficient of the x^2 term.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 25th 2010, 04:53 PM
  2. Quadratic Equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 18th 2010, 03:58 AM
  3. Quadratic Equation Help!
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 23rd 2009, 03:43 AM
  4. Quadratic equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 19th 2008, 06:38 PM
  5. Quadratic Equation
    Posted in the Algebra Forum
    Replies: 7
    Last Post: May 22nd 2008, 05:34 PM

Search Tags


/mathhelpforum @mathhelpforum