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Thread: 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:

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

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

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

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

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

    $\displaystyle 5=x^2+6x+10$

    $\displaystyle x^2+6x+5=0$

    $\displaystyle (x+5)(x+1)=0$

    $\displaystyle \boxed{x=-5 \ \ or \ \ x=-1}$
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  3. #3
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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

    $\displaystyle x^2+81=x^2+6x+9$

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

    $\displaystyle 81=6x+9$

    Subtracting 9 from both sides, we get

    $\displaystyle 72=6x$

    And dividing both sides by 6, we arrive at

    $\displaystyle 12 = x$
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