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Explain the methods for solving quadratics.  Do you understand the method / concept of completing the square and why we use it?  Describes the roots of a quadratic equation.  in what type of work a are quadratic equation used?

Jun 4th, 2015

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More help this is the link.........................http://www.purplemath.com/modules/quadform.htm

The quadratic in the previous section's last example, "(x – 2)2  – 12", can be multiplied out and simplified to be "x2 – 4x – 8". But we would not have been able to solve the equation with the quadratic formatted this way because it doesn't factor and it isn't ready for square-rooting. The only reason we could solve it before was because they'd already put all the x stuff inside a square, so we could square-root both sides. So how do you go from a regular quadratic like "x2 – 4x – 8" to one that is ready to be square-rooted? We would have to "complete the square".

I have a lesson on solving quadratics by completing the square, which explains the steps and gives examples of this process. It also shows how the Quadratic Formula is generated by this process. So I'll just do just one example of the process in this lesson. If you need further instruction, read the lesson at the above hyperlink.

  • Use completing the square to solve x2 – 4x – 8 = 0.

    As noted above, this quadratic does not factor, so I can't solve the equation by factoring. And they haven't given me the quadratic in a form that is ready to square-root. But there is a way for me to manipulate the quadratic to put it into that form, and then solve. It works like this:

    First, I put the loose number on the other side of the equation:

      x2 – 4x – 8 = 0
      x2 – 4x = 8 Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

    Then I look at the coefficient of the x-term, which is –4 in this case. I take half of this number (including the sign), giving me –2. Then I square this value to get +4, and add this squared value to both sides of the equation:

      x2 – 4x + 4 = 8 + 4
      x2 – 4x + 4 = 12

    This process creates a quadratic that is a perfect square, and factoring gives me:

      (x – 2)2 = 12

    Now I can square-root both sides of the equation, simplify, and solve:

      (x – 2)2 = 12

      Then the solution is 

    Unless you're told that you have to use completing the square, you will probably not use this method, in actual practice, when solving quadratic equations. Either some other method (such as factoring) will be obvious and quicker, or else the Quadratic Formula (coming up next) will be easier to use. However, if your class covered completing the square, you should expect to be required to show that you can complete the square to solve a quadratic on the next test.

    Often, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. While factoring may not always be successful, the Quadratic Formula can always find the solution.

    The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve. The Quadratic Formula is derived from the process of completing the square, and is formally stated as:

       For ax2 + bx + c = 0, the value of x is given by:

    For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root. And it's a "2a" under there, not just a plain "2". Make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself up. Remember that "b2" means "the square of ALL of b, including its sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive.

    In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!

    Here are some examples of how the Quadratic Formula works:

    • Solve x2 + 3x – 4 = 0

      This quadratic happens to factor:

        x2 + 3x – 4 = (x + 4)(x – 1) = 0

      ...so I already know that the solutions are x = –4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution looks like this:

        Then, as expected, the solution is x = –4, x = 1.

      Suppose you have ax2 + bx + c = y, and you are told to plug zero in for y. The corresponding x-values are the x-intercepts of the graph. So solving ax2 + bx + c = 0 for x means, among other things, that you are trying to find x-intercepts. Since there were two solutions for x2 + 3x – 4 = 0, there must then be two x-intercepts on the graph. Graphing, we get the curve below:


        As you can see, the x-intercepts (the red dots above) match the solutions, crossing the x-axis at
        x = –4 and x = 1. This shows the connection between graphing and solving: When you are solving "(quadratic) = 0", you are finding the x-intercepts of the graph. This can be useful if you have a graphing calculator, because you can use the Quadratic Formula (when necessary) to solve a quadratic, and then use your graphing calculator to make sure that the displayed x-intercepts have the same decimal values as do the solutions provided by the Quadratic Formula.

        Note, however, that the calculator's display of the graph will probably have some pixel-related round-off error, so you'd be checking to see that the computed and graphed values were reasonably close; don't expect an exact match.  Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

        • Solve 2x2 – 4x – 3 = 0.  Round your answer to two decimal places, if necessary.

          There are no factors of (2)(–3) = –6 that add up to –4, so I know that this quadratic cannot be factored. I will apply the Quadratic Formula. In this case, a = 2, b = –4, and c = –3:

            Then the answer is x = –0.58, x = 2.58, rounded to two decimal places.

          Warning: The "solution" or "roots" or "zeroes" of a quadratic are usually required to be in the "exact" form of the answer. In the example above, the exact form is the one with the square roots of ten in it. You'll need to get a calculator approximation in order to graph the x-intercepts or to simplify the final answer in a word problem. But unless you have a good reason to think that the answer is supposed to be a rounded answer, always go with the exact form.

Please let me know if you need any clarification. I'm always happy to answer your questions.
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