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Solving Equations by Factoring - Find the length of the hypotenuse

  1. Feb 26, 2009 #1
    "Solving Equations by Factoring" - Find the length of the hypotenuse

    1. The problem statement, all variables and given/known data

    One leg of a right triangle is 2 feet more than twice the other leg. The hypotenuse is 1 foot more than the longer leg. Find the length of the hypotenuse of the right triangle.

    * I am not sure how to calculate the square of the hypotenuse [tex](2n+2)+1 [/tex]

    Whether it's supposed to be [tex]4n^2+4+1 = 4n^2+5[/tex]
    or [tex]4n^2+9[/tex]

    Maybe this is why I don't know how to get the right answer?

    Given formulas:

    Hypotenuse denoted by c = [tex](2n+2)+1[/tex]
    Leg 1 denoted by a = [tex]2n+2[/tex]
    Leg 2 denoted by n = n

    The book's answer: The length of the hypotenuse is 13 feet.

    3. The attempt at a solution

    Attempt 1:

    [tex]c^2-a^2=n^2[/tex]
    [tex](4n^2+9)-(4n^2+4)=n^2[/tex]
    [tex]5=n^2[/tex]
    [tex]\sqrt{5}=n[/tex]
    [tex]n=2.2[/tex] I don't think they want decimals so I tried another way. Also, if I were to use [text] c=4n^2+5[/text] I would get n=1 again.

    Attempt 2:
    [tex]c^2=a^2+b^2[/tex]
    [tex](4n^2+4)+1=(4n^2+4)+(n^2)[/tex]
    [tex](4n^2+4)+1=5n^2+4[/tex]
    [tex](-n^2)+1=0[/tex]
    [tex]-(n^2+1)=0[/tex]
    [tex]n^2-1=0[/tex]
    [tex]n^2=1[/tex]
    [tex]n=1[/tex]

    Are there any hints that would help me a lot? Thanks folks.
     
    Last edited: Feb 26, 2009
  2. jcsd
  3. Feb 26, 2009 #2

    gabbagabbahey

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    Re: "Solving Equations by Factoring" - Find the length of the hypotenuse

    For starters, the length of the hypotenuse ca be simplified: [itex](2n+2)+1=2n+3[/itex], so it's square is just [itex](2n+3)^2[/itex]....how do you square any binomial?:wink:
     
  4. Feb 26, 2009 #3
    Re: "Solving Equations by Factoring" - Find the length of the hypotenuse

    Try it like this,

    Short Leg is x,

    Longer leg is 2x+2,

    Hypotenuse is 2x+3,

    Pythagorean theorem says that the sum of the squares of the two legs is equal to the square of the hypotenuse.

    So (x)^2 + (2x+2)^2 = (2x+3)^2. Try to solve that.

    Remember that (a+b)^2 is not equal to a^2 + b^2. You have to use foil when their is a plus sign in there.
     
  5. Feb 27, 2009 #4
    Re: "Solving Equations by Factoring" - Find the length of the hypotenuse

    I got confused because bedmas said brackets and exponents are first... i dunno
     
    Last edited: Feb 27, 2009
  6. Feb 28, 2009 #5
    Re: "Solving Equations by Factoring" - Find the length of the hypotenuse

    oh ok... now I know what you were saying, I got messed up and got confused about how you simplify exponents like [tex] (m^2m^7)^3=m^6m^{21}[/tex] and [tex](m+n)^2\neq m^2+n^2[/tex]
     
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