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Thread: How did they get this answer?

  1. #1
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    How did they get this answer?

    The equation is: 8x^2 - 3150x - 360000 = 0

    the answer is: x= 486.285

    i tried to solve this problem using the quadratic equation but I get a negative # in the "sqrt of b^2 - 4ac" and my calculator gives me an error. Can I not use the quadratic equation to solve for x? How do you get this answer?

    thanks
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  2. #2
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    Re: How did they get this answer?

    The equation is: 8x^2 - 3150x - 360000 = 0

    the answer is: x= 486.285
    I agree that "their" positive solution is close ...

    $b^2-4ac = (-3150)^2 - 4(8)(-360000) = 21442500$

    $x = \dfrac{3150 \pm \sqrt{21442500}}{16}$

    $x \approx 486.2877945$

    also, $x \approx -92.53779451$
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  3. #3
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    Re: How did they get this answer?

    Quote Originally Posted by esinc52 View Post
    The equation is: 8x^2 - 3150x - 360000 = 0

    the answer is: x= 486.285

    i tried to solve this problem using the quadratic equation but I get a negative # in the "sqrt of b^2 - 4ac" and my calculator gives me an error. Can I not use the quadratic equation to solve for x? How do you get this answer?

    thanks
    The answer you have given is correct. (There is a second answer which is negative and so may not be valid in the context of the original problem.)

    a=8, b=-3150, c=-360000

     b^2 - 4ac = (-3150)^2 - 4*8*(-360000) is positive.

    I'd say you have made the very common mistake of calculating b^2 = -3150^2 rather than (-3150)^2.
    Thanks from romsek
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  4. #4
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    Re: How did they get this answer?

    $8x^2 - 3150x - 360000 = 0$

    $a=8$
    $b=-3150$
    $c=-360000$

    $r_{1,2} = \dfrac{3150 \pm \sqrt{(-3150)^2 -(4)(8)(-360000)}}{2(8)} = \dfrac{3150 \pm \sqrt{21442500}}{16} = (486.288, -92.5378)$
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  5. #5
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    Re: How did they get this answer?

    Thank you for your help guys. I see where my mistake was. "c" should be -360000, and not 36000. Also "b" should be -3150, and not 3150. I was thinking that the "b" and "c" were still the same whether it was "ax^2 + bx + c = 0" or "ax^2 - bx - c = 0"
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