# Roots Problem!

• Mar 12th 2011, 11:31 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

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

Here is my thinking here.

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

• Mar 12th 2011, 11:35 PM
Prove It
What is $\displaystyle (x + y)^2$ when expanded?
• Mar 13th 2011, 12:08 AM
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

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

Here is my thinking here.

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

So you would have $\sqrt{1^2 + 1^2} = 1 + 1$ ....??!!
• Mar 13th 2011, 12:14 AM
Astar
Quote:

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

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

So therefore you can say that:

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

Then what does $\sqrt{x^2+y^2}$ equal then???
• Mar 13th 2011, 12:19 AM
CaptainBlack
Quote:

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

So therefore you can say that:

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

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

There is no general further simplification

CB
• Mar 13th 2011, 04:26 AM
iBurger
Quote:

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

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

What you want to do,

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

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