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Math Help - algebra

  1. #1
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    algebra

    Can some one please help me figure out how to solve this equation
    3x^2-5x-12=0
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  2. #2
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    Quote Originally Posted by imbadatmath View Post
    Can some one please help me figure out how to solve this equation
    3x^2-5x-12=0
    What have you learned in school so far on this?

    There is the quadratic method.

    There is factoring a trinomial in quadratic form into the product of 2 binomials.
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  3. #3
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    i think we have learned both ways but i am unsure how to do it both ways
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  4. #4
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    Quote Originally Posted by imbadatmath View Post
    i think we have learned both ways but i am unsure how to do it both ways
    Let's do the quadratic method first:

    The first step is to get your equation into quadratic form which is;
    ax^2 + bx + c = 0

    You have already done that. Next, we need to find our inputs to the equation. For the equation you entered of 3x^2-5x-12=0, we see that a = 3, b = -5, c = -12

    Read this writeup first: Quadratic equation - Wikipedia, the free encyclopedia

    Try doing that method. Then go here to check your answer and the math:

    Quadratic Equation

    Enter a,b,c values and press calculate. Memorize the formula and how to solve for the roots.

    When you are ready to learn the factoring method or if you have any question on the quadratic method, let me know.
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  5. #5
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    Factoring takes practice and at first it may seem difficult but once you get the method down every problem is essentially the same. What I do is first I write down two sets of parentheses:

    (____)(____)

    Then I look at the x^2 term and decide what two x terms can I multiply together to get the x^2 term. In this case the coefficient on the x^2 term is 3 so I need to have a 3x times an x since 3x*x = 3x^2. Write those in the first position.

    (3x___)(x___)

    Now I need to multiply two numbers together to get the x^0 term (the term within an x on it) in your case it is -12. Now there are several pairs of numbers I can multiply together to get -12. They are: -6*2, -2*6, -12*1, -1*12, -3*4, -4*3. Not just any of these numbers will work though. We need to choose the pair that when multiplied by the correct x-terms and added together give us the middle term in our quadratic (in your case the -5x).

    Lets try a few:
    first try, -4*3.
    remember FOIL (first outer inner last when multiplying out two binomials)
    (3x - 4)(x + 3) = 3x^2 +9x - 4x -12 = 3x^2 +5x -12.

    That is pretty close, the problem was that instead of -5x our middle term is +5x. How can we fix this? Instead of -4, and 3 lets try -3 and 4.

    (3x + 4)(x-3) = 3x^2 - 9x + 4x -12 = 3x^2 -5x - 12

    It worked!

    Now we set 3x+4 = 0 and x-3 = 0 and solve for x in each equation.
    _____________________
    It can get more complicated because maybe we have 6x^2-2x-1. In this case we could choose 3x and 2x or 6x andx for the first values in our binomials since when we multiply them together both come out to 6x^2.

    With practice you will be able to see right away which one is more likely to work and you won't usually have to try every possibility. Mathceleb might have some other tricks to factoring faster. Just remember it is like FOILing in reverse.
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  6. #6
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    so the 2 x values would be as followed?
    3x+4=0
    -4 -4
    3x=-4 / both sides by 3
    x=-1.33333333

    and

    x-3=0
    +3 +3
    x=3
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  7. #7
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    Exactly correct.
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  8. #8
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    Quote Originally Posted by imbadatmath View Post
    Can some one please help me figure out how to solve this equation
    3x^2-5x-12=0
    You could also "complete the square:"
    3x^2 - 5x - 12 = 0

    Isolate the quadratic and linear terms:
    3x^2 - 5x = 12

    Factor so that you have a coefficient of 1 on the quadratic term:
    3 \left ( x^2 - \frac{5}{3}x \right ) = 12

    The expression in the parenthesis is now in the form of x^2 + 2ax. If we add a^2 to this the term becomes a perfect square. And what we add to one side we must also subtract so we aren't changing anything:
    3 \left ( x^2 - \frac{5}{3}x + \frac{5^2}{6^2} - \frac{5^2}{6^2} \right ) = 12

    Separate out the first three terms:
    3 \left ( x^2 - \frac{5}{3}x + \frac{5^2}{6^2} \right ) - 3 \cdot \frac{5^2}{6^2} = 12

    Now solve for x:
    3 \left ( x - \frac{5}{6} \right ) ^2 - \frac{125}{36} = 12

    3 \left ( x - \frac{5}{6} \right ) ^2 = 12 + \frac{125}{36}

    3 \left ( x - \frac{5}{6} \right ) ^2 = \frac{169}{12}

    \left ( x - \frac{5}{6} \right ) ^2 = \frac{169}{36}

    x - \frac{5}{6} = \pm \sqrt{\frac{169}{36}}

    x = \frac{5}{6} \pm \frac{13}{6}

    So x = 3 or x = -\frac{4}{3}

    -Dan
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  9. #9
    Forum Admin topsquark's Avatar
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    There is another method that only works if the quadratic can be factored. It is called the "ac" method.
    3x^2 - 5x - 12 = 0

    Multiply the 3 and the -12 (the "a" and "c" coefficients of the quadratic):
    3 \cdot (-12) = -36

    Now list all the pairs of factors of -36:
    1, -36
    2, -18
    3, -12
    4, -9
    6, -6
    9, -4
    12, -3
    18, -2
    36, -1

    Now, if this quadratic factors there will be a pair of these that adds up to be the "b" coefficient, in this case -5. And in fact, there is: (4) + (-9) = -5.

    So rewrite the quadratic using the -5 as 4 + -9:
    3x^2 - 5x - 12 = 0

    3x^2 + (4 - 9)x - 12 = 0

    3x^2 + 4x - 9x - 12 = 0

    Now you want to "factor by grouping." Factor the first two terms together, and the last two terms together:
    (3x^2 + 4x) + (-9x - 12) = 0

    x(3x + 4) - 3(3x + 4) = 0

    Now you have a common factor (3x + 4) on both terms, so you can factor this:
    (x - 3)(3x + 4)

    Now solve normally and get
    x = 3 and x = -\frac{4}{3}

    -Dan
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