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Math Help - More Algebra

  1. #1
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    More Algebra

    A city lot has the shape of a right triangle whose hypotenuse is 7 ft longer than one of the other sides. The perimeter of the lot is 392 ft. How long is each side of the lot?
    Let x be one side, ? be another, and x + 7 be the hypotenuse. You find that ? = \sqrt{14x+49}. Plug all 3 into the perimeter formula with 392 as the perimeter.

    After some math, I got 14x + 49 = 4x^2 - 1540x + 148225. Subtract 14x + 49 and I get 0=4x^2-1554x+148176

    At this point I'm lost because the numbers I'm getting after plugging it into the quadratic equation are anywhere between 80-90% of the perimeter, which doesn't make sense. If anyone could help, that'd be great.
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  2. #2
    Eater of Worlds
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    The length is x^{2}+y^{2}=(x+7)^{2}

    The perimeter is x+y+x+7=2x+y+7=392

    y=385-2x

    Sub into the area formula:

    x^{2}+(385-2x)^{2}=(x+7)^{2}. What you have.

    Solve for x and we get x=168 and x=220.5

    The 168 is the good answer. Because 385-2(220.5) would result in a negative side length.

    So, we have side lengths of 168 and 49 and 175
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  3. #3
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    Your work is fine. You could have used the quadratic formula to find the roots.

    4x^2 - 1554x + 148176 = 0

    a = 4, b = -1554, c = 148176

    x = \frac{1554\pm\sqrt{(-1554)^2 - 4(4)(148176)}}{2(4)}

    x = \frac{1554\pm\sqrt{44,100}}{8}

    x = \frac{1554\pm210}{8}

    x = 220.5\ or\ x = 168

    Only of the the roots gives the correct answer.

    EDIT: aaah, galactus you beat me.
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  4. #4
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    It's cool, I'll give you both thanks.

    I don't know where I was getting the ridiculous numbers from. Maybe I did something wrong in my work. Not sure.
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