SOLUTION: please solve this sum :- if two sides are given 2x(x+1) and 2x+1, then what is the length of hypotenuse side of right angle triangle.

Algebra ->  Pythagorean-theorem -> SOLUTION: please solve this sum :- if two sides are given 2x(x+1) and 2x+1, then what is the length of hypotenuse side of right angle triangle.       Log On


   



Question 957249: please solve this sum :- if two sides are given 2x(x+1) and 2x+1, then what is the length of hypotenuse side of right angle triangle.

Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
use Pythagorean theorem to find the length of hypotenuse c of right angle triangle:
c%5E2=a%5E2%2Bb%5E2
given: two sides
a=+2x%28x%2B1%29 and b=2x%2B1
a=+2x%5E2%2B2x
c%5E2=%282x%5E2%2B2x%29%5E2%2B%282x%2B1%29%5E2
c%5E2=4x%5E4%2B8x%5E3%2B4x%5E2%2B4x%5E2%2B4x%2B1
c%5E2=4x%5E4%2B8x%5E3%2B8x%5E2%2B4x%2B1
c%5E2=%282x%5E2%2B2x%2B1%29%5E2
c=sqrt%28%282x%5E2%2B2x%2B1%29%5E2%29
c=2x%5E2%2B2x%2B1
so, the length of hypotenuse is highlight%282x%5E2%2B2x%2B1%29


Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
She's right BUT!!! -- how in the lulu was she 
able to go from

this step:

c%5E2=4x%5E4%2B8x%5E3%2B8x%5E2%2B4x%2B1

to this step????:

c%5E2=%282x%5E2%2B2x%2B1%29%5E2

It is correct but she doesn't explain how she did that.
I have no idea how she did that myself!  A magic wand!  :)

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I would not have multiplied everything out as she did.
Here's what I would have done:

c%5E2=%282x%28x%2B1%29%5E%22%22%29%5E2%2B%282x%2B1%29%5E2

I'd multiply %282x%2B1%29%5E2 out as 4x%5E2%2B4x%2B1

c%5E2=%282x%28x%2B1%29%5E%22%22%29%5E2%2B4x%5E2%2B4x%2B1

Then I'd realize that if we factor 4x out of 4x%5E2%2B4x
we get 4x%28x%2B1%29 which is 2 times the 2x(x+1) in the 
parentheses in the first term on the right:

c%5E2=%282x%28x%2B1%29%5E%22%22%29%5E2%2B4x%28x%2B1%29%2B1

Now to make things simpler, I will let u = 2x(x+1)
then 4x(x+1) will equal to 2u, and we have:



Taking positive square roots

c=u%2B1

And since u+=+2x%28x%2B1%29, we have:

c+=+2x%28x%2B1%29%2B1

c+=+2x%5E2%2B2x%2B1

Edwin