# Roots Problem!

• Mar 12th 2011, 10:31 PM
Astar
Roots Problem!
You know those times when your doing maths and then for some reason you just CANNOT work out why a particular mathematical situation is that way! Well I just did.

FOr the life of me I cannot work out why the following happens:

How come

$\displaystyle \sqrt {x^2+y^2}\not= x + y$

Here is my thinking here.

$\displaystyle \sqrt {x^2+y^2} = \sqrt {x^2} + \sqrt {y^2} = x + y$

• Mar 12th 2011, 10:35 PM
Prove It
What is $\displaystyle \displaystyle (x + y)^2$ when expanded?
• Mar 12th 2011, 11:08 PM
mr fantastic
Quote:

Originally Posted by Astar
You know those times when your doing maths and then for some reason you just CANNOT work out why a particular mathematical situation is that way! Well I just did.

FOr the life of me I cannot work out why the following happens:

How come

$\displaystyle \sqrt {x^2+y^2}\not= x + y$

Here is my thinking here.

$\displaystyle \sqrt {x^2+y^2} = \sqrt {x^2} + \sqrt {y^2} = x + y$

So you would have $\displaystyle \sqrt{1^2 + 1^2} = 1 + 1$ ....??!!
• Mar 12th 2011, 11:14 PM
Astar
Quote:

Originally Posted by Prove It
What is $\displaystyle \displaystyle (x + y)^2$ when expanded?

It equals $\displaystyle x^2 + 2xy + y^2$

So therefore you can say that:

$\displaystyle x+y = \sqrt{(x+y)^2}$

Then what does $\displaystyle \sqrt{x^2+y^2}$ equal then???
• Mar 12th 2011, 11:19 PM
CaptainBlack
Quote:

Originally Posted by Astar
It equals $\displaystyle x^2 + 2xy + y^2$

So therefore you can say that:

$\displaystyle x+y = \sqrt{(x+y)^2}$

Then what does $\displaystyle \sqrt{x^2+y^2}$ equal then???

There is no general further simplification

CB
• Mar 13th 2011, 03:26 AM
iBurger
Quote:

Originally Posted by Astar
...
Then what does $\displaystyle \sqrt{x^2+y^2}$ equal then???
...

I think it helps to read $\displaystyle \sqrt{x^2+y^2} = (x^2+y^2)^{.5} = (constant)^{.5}$

What you want to do,

$\displaystyle \sqrt {x^2+y^2} = \sqrt {x^2} + \sqrt {y^2} = x + y$

Would be almost correct if it was $\displaystyle \sqrt {x^2*y^2}$ instead of $\displaystyle \sqrt {x^2+y^2}.$