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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.

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# Homework Help: Solving Equations by Factoring - Find the length of the hypotenuse

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