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Thread: Solve the following equation graphically to two decimal places using a graphing cal..

  1. #1
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    Solve the following equation graphically to two decimal places using a graphing cal..

    1.2 x^2 - 1.3 x - 5.5

    Which values of x, to two decimal places, satisfy the inequality? The answer is -1.67<x<2.75 How do I solve this?

    Could I use a graphing calculator so I can solve for these? I must have two answers.
    Last edited by TexasGuy09; Sep 13th 2017 at 05:57 PM.
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  2. #2
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    Re: Solve the following equation graphically to two decimal places using a graphing c

    Quote Originally Posted by TexasGuy09 View Post
    1.2 x^2 - 1.3 x - 5.5 ...

    Which values of x, to two decimal places, satisfy the inequality? The answer is -1.67<x<2.75 How do I solve this?

    Could I use a graphing calculator so I can solve for these? I must have two answers.
    What inequality?. All I see is a quadratic expression ...
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  3. #3
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    Re: Solve the following equation graphically to two decimal places using a graphing c

    I must solve for the two x's.
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    Re: Solve the following equation graphically to two decimal places using a graphing c

    Never mind. I discovered a method. :0
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    Re: Solve the following equation graphically to two decimal places using a graphing c

    Don't worry. I discovered this solution myself!
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    Re: Solve the following equation graphically to two decimal places using a graphing c

    But we still don't know what the problem was! And we are curious.
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    Re: Solve the following equation graphically to two decimal places using a graphing c

    It was probably "find all x for which $\displaystyle 1.2x^2- 1.3x- 5.5< 0$". Using a graphing calculator, you graph $\displaystyle y= 1.2x^2- 1.3x- 5.5$ (I used the one at https://www.desmos.com/calculator). Notice that parabola goes below the x-axis and, if necessary use the zoom feature to determine that the parabola crosses the x-axis at "x= -1.667" (which is probably -5/3) and at "x= 2.75" (11/4). We can check that "probably" by setting x= -5/3 in the inequality- 1.2(25/9)- 1.3(-5/3)- 5.5= 0.

    Or you could use the quadratic formula to determine the endpoints of the interval: $\displaystyle x= \frac{1.3\pm\sqrt{1.3^2- 4(1.2)(5.5)}}{2(1.2)}= \frac{1.3\pm\sqrt{28.09}}{2.4}= \frac{1.3\pm 5.3}{2.4}$.
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