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 $$(2n+2)+1$$

Whether it's supposed to be $$4n^2+4+1 = 4n^2+5$$
or $$4n^2+9$$

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

Given formulas:

Hypotenuse denoted by c = $$(2n+2)+1$$
Leg 1 denoted by a = $$2n+2$$
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:

$$c^2-a^2=n^2$$
$$(4n^2+9)-(4n^2+4)=n^2$$
$$5=n^2$$
$$\sqrt{5}=n$$
$$n=2.2$$ 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:
$$c^2=a^2+b^2$$
$$(4n^2+4)+1=(4n^2+4)+(n^2)$$
$$(4n^2+4)+1=5n^2+4$$
$$(-n^2)+1=0$$
$$-(n^2+1)=0$$
$$n^2-1=0$$
$$n^2=1$$
$$n=1$$

Are there any hints that would help me a lot? Thanks folks.

 

[gabbagabbahey]

For starters, the length of the hypotenuse ca be simplified: $(2n+2)+1=2n+3$, so it's square is just $(2n+3)^2$...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: $(2n+2)+1=2n+3$, so it's square is just $(2n+3)^2$...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 $$(m^2m^7)^3=m^6m^{21}$$ and $$(m+n)^2\neq m^2+n^2$$