Square Root Rules: x^(1/2), abs(x), F(x), & More!

[Emmanuel_Euler]

hi guys
i know all square root and any root(cubic...) rules

sqrt(x)=x^(1/2)
sqrt(x^2)=abs(x)
sqrt(xy)=sqrt(x)*sqrt(y)
sqrt(x/y)=sqrt(x)/sqrt(y)
sqrt(-x)=isqrt(x)
f'(x)=1/2sqrt(x)
F(x)=2/3*(x^3/2)
.....
my question is:
is there any rules for this sqrt(x+y)
or sqrd(x-y)??

any help please??

 

[HallsofIvy]

No, there is no way to simplify a square root (or other root) of a sum or difference.

It's simply a case of "multiplication and addition do not play well together!"

 

[Emmanuel_Euler]

Yes of course.
Yesterday i was searching about this(sqrt(x+y)),i found nothing.
you are right there is no rule for this sqrt(x+y)
thanks for help friend.

 

[Mark44]

hi guys
i know all square root and any root(cubic...) rules

sqrt(x)=x^(1/2)
sqrt(x^2)=abs(x)
sqrt(xy)=sqrt(x)*sqrt(y)

You should be aware that there are restrictions on x and y; namely, both must be nonnegative. I.e., x ≥ 0 and y ≥ 0. Without these restrictions you get nonsense like $1 = \sqrt{1} = \sqrt{-1 * -1} = \sqrt{-1} * \sqrt{-1} = i * i = -1$

sqrt(x/y)=sqrt(x)/sqrt(y)

There are restrictions here, as well, with x ≥ 0 and y > 0.

sqrt(-x)=isqrt(x)

Not true. For example, if x = -4, then $\sqrt{-(-4)} = \sqrt{4} = 2$. $i\sqrt{-4} = i * (2i) = 2i^2 = -2$.
Here you seem to be tacitly assuming that -x will be negative, which is not true in general.

f'(x)=1/2sqrt(x)
F(x)=2/3*(x^3/2)

.....
my question is:
is there any rules for this sqrt(x+y)
or sqrd(x-y)??

any help please??

 

[Emmanuel_Euler]

You should be aware that there are restrictions on x and y; namely, both must be nonnegative. I.e., x ≥ 0 and y ≥ 0. Without these restrictions you get nonsense like $1 = \sqrt{1} = \sqrt{-1 * -1} = \sqrt{-1} * \sqrt{-1} = i * i = -1$
There are restrictions here, as well, with x ≥ 0 and y > 0.
Not true. For example, if x = -4, then $\sqrt{-(-4)} = \sqrt{4} = 2$. $i\sqrt{-4} = i * (2i) = 2i^2 = -2$.
Here you seem to be tacitly assuming that -x will be negative, which is not true in general.

Friend:i know all the rules you wrote.
i was too busy to write them all(in my question).
But if you really want to help me,find a rule for this sqrt(x+y).

 

[Emmanuel_Euler]

You should be aware that there are restrictions on x and y; namely, both must be nonnegative. I.e., x ≥ 0 and y ≥ 0. Without these restrictions you get nonsense like $1 = \sqrt{1} = \sqrt{-1 * -1} = \sqrt{-1} * \sqrt{-1} = i * i = -1$
There are restrictions here, as well, with x ≥ 0 and y > 0.
Not true. For example, if x = -4, then $\sqrt{-(-4)} = \sqrt{4} = 2$. $i\sqrt{-4} = i * (2i) = 2i^2 = -2$.
Here you seem to be tacitly assuming that -x will be negative, which is not true in general.

Forgive me!, i was busy and hurry.

 

[Mark44]

Friend:i know all the rules you wrote.
i was too busy to write them all(in my question).
But if you really want to help me,find a rule for this sqrt(x+y).

There is no such rule.

Period.

 

[Emmanuel_Euler]

You are right.

 

[micromass]

If you accept infinite sums, then there is always the binomial formula: http://en.wikipedia.org/wiki/Binomial_series

 

[Emmanuel_Euler]

That helps.
thank you for help,i will read it later!