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Math Help - Solving for x & side length in a right triangle

  1. #1
    VAP
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    Solving for x & side length in a right triangle

    Hello,
    I have to solve for x and side lengths in a right triangle.
    1. The two equidistant sides are x and x + 6
    2. The hypotenuse is 2 sqrt 17

    If I am correct so far, I used the Pythagorean Theorem. So:

    x^2 + (x + 6)^2 = (2 sqrt 17)^2

    x^2 + x^2 + 36 = I don't know how to do this.

    Thank you very much!
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  2. #2
    o_O
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    Important thing to note. In general: (a+b)^{2}\: {\color{red}\neq} \: a^{2} + b^{2}

    You have to multiply it out with the FOIL method you've learned at school: (a+b)^{2} = (a+b)(a+b) = ... = a^{2} + 2ab + b^{2}

    Recall that: (ab)^{c} = a^{c}b^{c}

    So for your radical term, imagine a = 2 and b = sqrt{17}.
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by VAP View Post
    Hello,
    I have to solve for x and side lengths in a right triangle.
    1. The two equidistant sides are x and x + 6
    2. The hypotenuse is 2 sqrt 17

    If I am correct so far, I used the Pythagorean Theorem. So:

    x^2 + (x + 6)^2 = (2 sqrt 17)^2

    x^2 + x^2 + 36 = I don't know how to do this.

    Thank you very much!
    The problem is :

    (x+6)^2=x^2+2 \cdot 6 \cdot x+36

    (2 \sqrt{17})^2=2^2 \cdot (\sqrt{17})^2=4 \cdot 17=68


    Therefore :

    x^2+x^2+12x+36=68

    2x^2+12x-32=0

    Divide by 2 :

    x^2+6x-16=0

    Complete the square :

    x^2+6x+3^2-3^2-16=0

    (x+3)^2-25=(x+3)^2-5^2=0

    Using the identity : a^2-b^2=(a-b)(a+b) :

    (x+3-5)(x+3+5)=0


    Continue... or finish
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by Moo View Post
    (x+6)^2=x^2+2 \cdot 6 \cdot x+36

    (2 \sqrt{17})^2=2^2 \cdot (\sqrt{17})^2=4 \cdot 17=68


    Therefore :

    x^2+x^2+12x+36=68

    2x^2+12x-32=0

    Divide by 2 :

    x^2+6x-16=0

    Complete the square :

    x^2+6x+3^2-3^2-16=0

    (x+3)^2-25=(x+3)^2-5^2=0

    Using the identity : a^2-b^2=(a-b)(a+b) :

    (x+3-5)(x+3+5)=0


    Continue... or finish
    -----------------------------------
    Moo, I liked your approach to this.

    I would just say that when you get to x^2+6x-16=0,
    try to factor the trinomial either "guess and check" or this way:

    Ask yourself, 'what 2 factors when multiplied together give you -16 and when added give you +6 (the coef. of the middle term)?'

    Not many factors of 16 to worry with, so let's try +8 and -2. I believe that'll do the trick.

    (x+8)(x-2)=0

    The zero product property tell us that x+8=0, or x-2 - 0

    Then, x=-8, or x=2

    x=-8 is extraneoous since the lengths cannot be negative.

    Therefore, x=2 and x+6=8
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  5. #5
    Moo
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    Hi masters

    Actually, I use this method for explaining to people... I would use what you've described, or just calculating the discriminant quickly to see what the solutions look like, if I had to deal with the problem

    Thanks for noticing !


    (VAP : please send mails instead of PMs, it will be a pleasure )



    --- This is my 2000th post
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by Moo View Post
    Hi masters



    --- This is my 2000th post

    CONGRATULATIONS on a great milestone....one that I may never reach. Keep up the great postings.

    Dale
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