Results 1 to 6 of 6

Math Help - Quadratic Equation Help please.

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    4

    Quadratic Equation Help please.

    Hi,

    I need help with this equation, i must solve it by factorising this quadratic equation.

    27x^2 - 2x = 8

    Help appreciated.

    Tim
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,860
    Thanks
    742
    Hello, Tim!

    Is there a typo?


    Solve it by factoring: . 27x^2 - 2x \:= \:8
    The quadratic: . 27x^2 - 2x - 8 \:=\:0 cannot be factored.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    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 Soroban,

    Why ?

    The discriminant is positive... Sure, the solution is quite ugly, but it remains factorisable.
    Well this is certainly not the way to do it since it's asked to factorize first but it's false to state what you said...

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,860
    Thanks
    742
    Hello, Moo!

    The discriminant is positive...
    Sure, the solution is quite ugly, but it remains factorisable.
    As soon as you mention "Discriminant", you are no longer factoring.
    . . You are obviously invoking the Quadratic Formula.


    But it's false to state what you said...
    I don't agree . . .

    According to my understanding of "factorable", x^2-4x + 6 doesn't factor.

    Sure, we can use the Quadratic Formula and get: . x \;=\;2 \pm i\sqrt{5}

    We can say that we have two factors: . \left(x - [2 + i\sqrt{5}]\right)\,\left(x - [2-i\sqrt{5}]\right)

    . . but I would not claim that we factored it.


    If we include irrational and complex expressions, everything factors.

    . . x^4 + 1 \;=\;\left(x-\sqrt{i}\right)\left(x + \sqrt{i}\right)\left(x - i\sqrt{i}\right)\left(x + i\sqrt{i}\right)

    . . x + 4 \;=\;\left(\sqrt{x} - 2i\right)\left(\sqrt{x}+2i\right)

    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
    Oh, I think we just don't agree with the term "factorable"...
    There will always be a way to factor it

    But I'm not sure to get the reason why it is not factorable in your terms...
    Your example of x-4x+6 shows that we factored it... since we got two factors... whatever the way we used is, isn't it ?

    There is also the property :
    if a trinomial is factorable, then, it can be written the following way : c(x-a)(x-b).

    Huh ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member cassiopeia1289's Avatar
    Joined
    Aug 2007
    From
    chicago
    Posts
    101
    ok - back to the actual question - use your quadratic equation with a=27, b=-2, and c =-8
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 16th 2011, 08:03 AM
  2. Quadratic equation
    Posted in the Algebra Forum
    Replies: 5
    Last Post: March 16th 2011, 11:27 AM
  3. Replies: 3
    Last Post: April 25th 2010, 04:53 PM
  4. Quadratic Equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 14th 2009, 06:34 AM
  5. quadratic equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 15th 2008, 05:06 AM

Search Tags


/mathhelpforum @mathhelpforum