Results 1 to 3 of 3

Math Help - Solving quadratic equations

  1. #1
    Newbie
    Joined
    Feb 2007
    Posts
    6

    Solving quadratic equations

    This stuff is kicking my butt!!!
    1. 6x^2-7x-5=0

    Complete the square
    2.x^2-4x-18=0
    3.4x^2-12x-3=0
    Thanks for the help guys
    Follow Math Help Forum on Facebook and Google+

  2. #2
    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 ls1ramair19 View Post
    1. 6x^2-7x-5=0
    Please do not double post the same question, it is against the rules.

    1. 6x^2 - 7x - 5 = 0
    You can use the quadratic formula on this and have it done in one step. Use the formula when you are not sure what to do and the method you have to do it by is not specified. (Youd do know the formula right?)

    If you don't want to use the formula, here is a standard method to do this problem besides completing the square.

    we multiply the first term, that is x^2 and its coefficient by the constant, so we obtain -30x^2. then we take the middle term, which is - 7x.

    Now we try to think of two numbers that multiplied gives -30x^2 and added gives -7x

    -30x^2 = 3x*(-10x)
    -7x = 3x + (-10x)

    so our numbers are 3x and -10x, now we replace the middle term with these two numbers in such a way that we can group the first two terms and the last two terms and factor out a common factor. so we rewrite the quadratic as

    6x^2 + 3x - 10x - 5 = 0
    => 3x(2x + 1) - 5(2x + 1) = 0 ............factored 3x out of the first two terms and -5 out of the last two.
    Now we see 2x + 1 is a common factor, so factor that out
    => (2x + 1)(3x - 5) = 0
    => 2x + 1 = 0 or 3x - 5 = 0
    => x = -1/2 or x = 5/3

    Complete the square
    3.4x^2-12x-3=0
    I want you to try the 2nd one on your own, it is simpler than this one. Just follow my steps.

    Here is completing the square. The first thing we want to do is make sure the coefficient of x^2 is 1

    4x^2 - 12x - 3 = 0
    => x^2 - 3x - 3/4 = 0 .............divided through by 4 to get the coefficient of x^2 as 1

    Now move the constant over the other side

    => x^2 - 3x = 3/4

    Now to have a complete square, the constant we have must be the square of half the coefficient of x, so let's add it. we have to add it to both sides to maintain the equality. the coefficient of x is -3, half of -3 is -3/2

    => x^2 - 3x + (-3/2)^2 = 3/4 + (-3/2)^2

    Now we contract the left hand side, using x and the new constant we added and square it

    => (x - 3/2)^2 =
    3/4 + (-3/2)^2
    => (x - 3/2)^2 = 3/4 + 9/4 = 3

    Now all that's left is to solve for x

    => (x - 3/2)^2 = 3
    => x - 3/2 = +/- sqrt(3) .............took the squareroot of both sides
    => x = 3/2 +/- sqrt(3)
    => x = 3/2 + sqrt(3) or x = 3/2 - sqrt(3)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Now try the 2nd one and post your solution
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving quadratic equations
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: November 10th 2009, 03:56 AM
  2. Solving Quadratic Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 31st 2009, 05:08 AM
  3. Solving Quadratic Equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 28th 2008, 11:48 AM
  4. Replies: 1
    Last Post: June 12th 2008, 10:30 PM
  5. Solving Quadratic Equations
    Posted in the Algebra Forum
    Replies: 12
    Last Post: January 9th 2008, 08:36 PM

Search Tags


/mathhelpforum @mathhelpforum