Solving Equations by Factoring - Find the length of the hypotenuse

[Raizy]

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

Homework Statement

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.

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.

 

[gabbagabbahey]

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?

 

[letmeknow]

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.

 

[Raizy]

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?

I got confused because bedmas said brackets and exponents are first... i dunno

 

[Raizy]

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]