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Thread: Finding the Solution in a Quadratic Equation?

  1. #1
    Feb 2010

    Finding the Solution in a Quadratic Equation?


    I'm back again with a few problems I'm stuck on. I'm really not sure where to go with them, or how to solve them. Unfortunately I'm a complete dunce when it comes to math, and only manage to get by with help from tutors and such. But these were a few problems that I didn't have time to ask at the tutor's...

    I don't know if it makes any difference, but here are some of the instructions that were printed at the top of my paper...

    Find the solution for each of the following equations.
    1- Put the equation in $\displaystyle Ax^2 + Bx + C = 0$ form
    2- Determine the form of the equation
    3- Find the solutions
    -Set each factor equal to 0 and solve
    -Solve using the square root method ( $\displaystyle \sqrt{Ax^2 } = +- \sqrt{C}$)
    -Solve by factoring
    -Use the Quadratic Equation

    Most of the problems are the $\displaystyle Ax^2 + Bx + C = 0$ form, and I know that if there is nothing that adds to B and multiplies to C (or maybe it's the other way around) that I need to use the square root method, which is different for me as before we were just told to write "no solution". This time I guess we have to continue it.

    I've got most of the problems solved, but there's still a few I'm struggling with and I'm not sure where to go with them. They're probably easier than I think they are, so I'm probably over-thinking things, but...

    1. $\displaystyle x^2 + 9x = 0$
    For this, do I just bring the 0 over and than add the 0 back in at the end as well, and than solve, or is it something different?
    Though I guess bringing over the 0 doesn't make much sense, now that I think about it...

    2. $\displaystyle 3x^2 = 60$
    I'm really not sure where to go with this one, as it has no Bx and such...

    3. $\displaystyle (3x-4)^2 = 16$

    4. $\displaystyle 4x^2 - 16 = 0$

    Also, for a couple problems I finished I'm just not sure if I got the correct answer. Would anyone be willing to just look them over for me and tell me whether I got them right or wrong?

    Problem: $\displaystyle 4x^2 = 2x + 7$

    Answer: $\displaystyle \frac {1 + \sqrt{29 }}{2}$ or $\displaystyle \frac {1 - \sqrt{29 }}{2}$

    Problem: $\displaystyle x^2 -2x - 10 = 0$

    Answer: $\displaystyle \frac {1 + \sqrt{11 }}{1}$ or $\displaystyle \frac {1 - \sqrt{11 }}{1}$

    Any help or advice would be very much appreciated. Thank you!
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  2. #2
    Super Member
    Jul 2009
    $\displaystyle x^{2}+9x=0 \Rightarrow x^{2}+9x+0=0 ; a=1, b=9, c=0$

    You can use the quadratic here, or complete the square and solve. Using quadratic:

    $\displaystyle x=\frac{-9\pm\sqrt{9^{2}-4(1)(0)}}{2(1)}$

    $\displaystyle 3x^2 = 60$

    You are half right; though the "Bx" isn't visible, it is there. Your B, as with your C in question one, is 0, therefore the term drops off. I could just as easily write this as:

    $\displaystyle 3x^2+0x = 60$

    Or we could use: "-Solve using the square root method"

    $\displaystyle \sqrt{3x^2 } = \pm \sqrt{60}$

    $\displaystyle 4x^2 = 2x + 7$
    $\displaystyle 4x^2-2x-7=0$

    Using the quadratic:
    $\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^{2}-4(4)(-7)}}{2(4)}$ - Set up our quadratic.

    $\displaystyle x=\frac{2\pm\sqrt{4+112}}{8}$ - Perform operations.

    $\displaystyle x=\frac{2\pm\sqrt{4(29)}}{8}$ - Factor a 4 out of the quantity under the square root sign.

    $\displaystyle x=\frac{2\pm 2\sqrt{29}}{8}$ - Take the square root of 4, and bring it out of the square root sign.

    $\displaystyle x=\frac{1\pm\sqrt{29}}{4}$ - Dive by 2.

    All of the problems you have can be solved using all off the methods you outlined in your problem - it is just easier to use one method over another. For example:

    $\displaystyle 3x^2+0x = 60$

    Can be solved using the quadratic, or much quicker to use the "square root method" as I mentioned. Your task is figuring out which is the easiest way to go about solving.

    To check that you understand how to do these problems, use multiple methods to solve the same problem.
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