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Math Help - Factoring Quadratic equations

  1. #1
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    Factoring Quadratic equations

    I remember doing something like this in Algebra 2, but I am in PreCal so sorry if it is the wrong section..

    The equation is:
    2x^2+9x+4=0

    When I plug this equation into the quadratic formula I get the same answers my teacher got: x=-1/2 & x=-4

    This is how she factored it:
    2x^2+9+4=0
    (2x+1)(x+4)=0
    2x+1=0 | x+4=0
    2x=-1 | x=-4
    x=-1/2

    The problem I have w/ understanding this equation is the numbers inside the parentheses. We were taught to find factors of 4 that add up to 9 and 4+1 clearly doesn't equal 9. Can someone please explain to me how she got those numbers?

    /first post. This looks like a cool site and I will be coming here for math help from now on. Hopefully I will be able to answer just as many questions as I propose
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  2. #2
    Super Member Aryth's Avatar
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    The two in the x^2 term is what throws everything off. Look at the equation with the two divided through it:

    x^2 + \frac{9}{2} + 2 = 0

    Now we need two numbers that add up to \frac{9}{2} and multiply to be 2. Since the second term has a half in it, that's probably going to be part of it, we can try \frac{1}{2}

    \left(x + \frac{1}{2}\right)(x + a) = 0

    Now, what number can we pick? We can solve:

    \frac{1}{2} + a = \frac{9}{2}

    If you work it out you get that:

    a = 4

    So:

    \left(x + \frac{1}{2}\right)(x + 4) = 0

    Is your answer, which is exactly what you and your teacher got, and even though it took a little bit of extra work, just understand that if there is a number infront of the x^2 term then at least ONE x will have to be multiplied by something. Dividing through helps sometimes, but doing it the way your teacher did will be important for larger numbers.
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  3. #3
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    Another method, which is practically the same but doesn't involve fractions, is to find two numbers that multiply to 8 (the 8 comes from multiplying the 4 and 2) and add to 9. 8 and 1 will do. Now break up the terms as 2x^2+9x+4=(2x^2+8x)+(x+4)=2x(x+4)+(x+4)=(2x+1)(x+4  ).
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  4. #4
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    Quote Originally Posted by SpencerMakesMath View Post
    I remember doing something like this in Algebra 2, but I am in PreCal so sorry if it is the wrong section..

    The equation is:
    2x^2+9x+4=0

    When I plug this equation into the quadratic formula I get the same answers my teacher got: x=-1/2 & x=-4

    This is how she factored it:
    2x^2+9+4=0
    (2x+1)(x+4)=0
    2x+1=0 | x+4=0
    2x=-1 | x=-4
    x=-1/2

    The problem I have w/ understanding this equation is the numbers inside the parentheses. We were taught to find factors of 4 that add up to 9 and 4+1 clearly doesn't equal 9. Can someone please explain to me how she got those numbers?
    "Finding factors of c that add up to b" only works for quadratics of the form x^2+ bx+ c- that is, where the coefficient of x^2 is 1. If the coefficient of x is not 1 then you have to take factors of that coefficient into account also. Here the coefficient of x^2 is 2, which has factors of 2 and 1 and 2(4)+ 1(1) is equal to 9.


    first post. This looks like a cool site and I will be coming here for math help from now on. Hopefully I will be able to answer just as many questions as I propose
    Follow Math Help Forum on Facebook and Google+

  5. #5
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    There's a trick I picked up back in precalc that might help you. There's a billion ways to factor (bottom's up, etc) but I find this one the most intuitive, and I have yet to find a factoring problem that it doesn't work on (that has factors...of course).

    Given:

    2x^2+9x+4=0


    Step 1:

    Multiple the 'C' term and the 'A' term together. In this example we'll get 8x^2.


    Step 2:

    Now we need to find factors of 8x^2 that add up to get 9x. Clearly, the only factors that will work here are 8 and 1.


    Step 3:

    Next, we need to rewrite the problem like so:

    (2x^2+8x)(x+4)

    How did we get these numbers? All I did was bring down the factors of 8x^2 and group them with the squared term 2x^2 and the constant 4.

    Step 4:

    Now we simplify. The idea is to get the simplest group of terms on both sides.

    2x(x+4) 1(x+4)

    I have a bad habit of forgetting the understood one (even in university level math!) so I always make sure to write it.


    Step 5:

    Simply combine the outside terms into a group, and use the (x+4) group as itself. With this we get:

    (2x+1)(x+4)


    Step 6:


    Set it equal to zero and solve algebraically:

    (2x+1)(x+4)=0
    2x+1=0
    x+4=0
    x = -4, 1/2


    I hope this helps! I don't remember it's name, but if you can somehow find the name there's a few videos on youtube about it so you can see it worked on paper (which makes it even easier).
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