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Math Help - Couple math problems

  1. #1
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    Couple math problems

    Need a little help with these two questions:


    1. Find x so that the distance between (-3,2) and (x,1) is the square root of 5.


    Next I had a triangle, needed to find x. Used the pythagorean theorem and got this:


    x^2 + 81 = x^2 + 6x + 9


    Not sure what to do there, but I know it's basic algebra.


    Any help is appreciated.
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  2. #2
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    Quote Originally Posted by amw9490 View Post
    Need a little help with these two questions:


    1. Find x so that the distance between (-3,2) and (x,1) is the square root of 5.


    Next I had a triangle, needed to find x. Used the pythagorean theorem and got this:


    x^2 + 81 = x^2 + 6x + 9


    Not sure what to do there, but I know it's basic algebra.


    Any help is appreciated.
    Hi amw9490,

    I'm sure you know the distance formula:

    d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

    Substitute your two points: (-3, 2) and (x, 1)

    \sqrt{5}=\sqrt{(x+3)^2+(1-2)^2}

    \sqrt{5}=\sqrt{x^2+6x+9+1}

    \sqrt{5}=\sqrt{x^2+6x+10}

    5=x^2+6x+10

    x^2+6x+5=0

    (x+5)(x+1)=0

    \boxed{x=-5 \ \ or \ \ x=-1}
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by amw9490 View Post

    Next I had a triangle, needed to find x. Used the pythagorean theorem and got this:


    x^2 + 81 = x^2 + 6x + 9


    Not sure what to do there, but I know it's basic algebra.


    Any help is appreciated.
    Assuming what you have done up to this equation is correct, we have

    x^2+81=x^2+6x+9

    If we subtract x^2 from both sides we end up with

    81=6x+9

    Subtracting 9 from both sides, we get

    72=6x

    And dividing both sides by 6, we arrive at

    12 = x
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