# Simplifying radical expressions with variables

• Oct 14th 2011, 04:13 AM
vaironxxrd
Hello forum vaironxxrd here.

I have this problem that is, a radical with two variables in it.

Problem: $\displaystyle \sqrt{25x^2y}$

Solution:$\displaystyle \sqrt{25x^2y}$ = $\displaystyle x\sqrt{5 \cdot 4 y}$
= $\displaystyle 2x\sqrt{5y}$

Is that solution right? or what am I doing wrong?
• Oct 14th 2011, 04:20 AM
Plato
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by vaironxxrd
Problem: $\displaystyle \sqrt{25x^2y}$
Solution:$\displaystyle \sqrt{25x^2y}$ = $\displaystyle x\sqrt{5 \cdot 4 y}$
= $\displaystyle 2x\sqrt{5y}$
Is that solution right? or what am I doing wrong?

Just about all of that is incorrect.
$\displaystyle \sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$
• Oct 14th 2011, 07:09 AM
e^(i*pi)
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by vaironxxrd
Hello forum vaironxxrd here.

I have this problem that is, a radical with two variables in it.

Problem: $\displaystyle \sqrt{25x^2y}$

Solution:$\displaystyle \sqrt{25x^2y}$ = $\displaystyle x\sqrt{5 \cdot 4 y}$
= $\displaystyle 2x\sqrt{5y}$

Is that solution right? or what am I doing wrong?

$\displaystyle 5 \times 4 \neq 25$

Indeed $\displaystyle 5 \times 5 = 25$ so you have $\displaystyle \sqrt{5^2x^2y}$
• Oct 14th 2011, 08:15 AM
hachataltoolimhakova
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by Plato
Just about all of that is incorrect.
$\displaystyle \sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$

Out of interest, why is the 5 not +-5? I see that x is the absolute value, is this because it was x^2 under the radical therefore 'originally' a + or - number, whereas 25 was, well, just 25?
Thanks.
• Oct 14th 2011, 08:37 AM
Plato
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by hachataltoolimhakova
Out of interest, why is the 5 not +-5? I see that x is the absolute value, is this because it was x^2 under the radical therefore 'originally' a + or - number, whereas 25 was, well, just 25?

Because $\displaystyle \left( {\forall x} \right)\left( {\forall y \geqslant 0} \right)\left[ {\sqrt {25x^2 y} \geqslant 0} \right]$
• Oct 14th 2011, 10:01 AM
hachataltoolimhakova
Re: Simplifying radical expressions with variables
Thanks for the response. I do not know what that means since I am not familiar with the notation. Am I correct in believing $\displaystyle (\ -5 \left \mid x \mid \left \sqrt{y})^2$ yields the same result as using 5? Again, thanks for the response. An explanation of that notation would be good. I found 'for all', that was all I could interpret.

EDIT: Ah, is it because it is the principle square root and since y is positive, 5 has to be positive? Otherwise the overall term would be negative, and it wouldn't be the prinicpal square root?
• Oct 14th 2011, 11:59 AM
HallsofIvy
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by hachataltoolimhakova
Thanks for the response. I do not know what that means since I am not familiar with the notation. Am I correct in believing $\displaystyle (\ -5 \left \mid x \mid \left \sqrt{y})^2$ yields the same result as using 5? Again, thanks for the response. An explanation of that notation would be good. I found 'for all', that was all I could interpret.

EDIT: Ah, is it because it is the principle square root and since y is positive, 5 has to be positive? Otherwise the overall term would be negative, and it wouldn't be the prinicpal square root?

Yes, that is true. While both $\displaystyle 2^2= 4$ and $\displaystyle (-2)^2= 4$. The square root function is defined by "$\displaystyle \sqrt{a}$ is the non-negative number whose square is a" in order that it be single valued.
• Oct 14th 2011, 02:08 PM
vaironxxrd
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by Plato
Just about all of that is incorrect.
$\displaystyle \sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$

Quote:

Originally Posted by e^(i*pi)
$\displaystyle 5 \times 4 \neq 25$

Indeed $\displaystyle 5 \times 5 = 25$ so you have $\displaystyle \sqrt{5^2x^2y}$

Thanks to everyone who gave a response.

I'm and should be ashamed of not checking my answer here is another problem to prove, I have corrected myself.

Problem: $\displaystyle \sqrt{90x^4y^2}$

Solution:$\displaystyle x^2y\sqrt{90}$ = $\displaystyle x^2y\sqrt{3 \cdot 3 \cdot 5 \cdot 2}$ =

$\displaystyle 3x^2y\sqrt{5 \cdot 2}$ = $\displaystyle 3x^2y\sqrt{10}$
• Oct 14th 2011, 02:18 PM
Plato
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by vaironxxrd
Problem: $\displaystyle \sqrt{90x^4y^2}$
Solution:$\displaystyle 3x^2{\color{red}y}\sqrt{10}$

NO it must be $\displaystyle 3x^2|y|\sqrt{10}$
You see that $\displaystyle y$ could be negative.
• Oct 14th 2011, 02:34 PM
vaironxxrd
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by Plato
Just about all of that is incorrect.
$\displaystyle \sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$

Quote:

Originally Posted by e^(i*pi)
$\displaystyle 5 \times 4 \neq 25$

Indeed $\displaystyle 5 \times 5 = 25$ so you have $\displaystyle \sqrt{5^2x^2y}$

Quote:

Originally Posted by Plato
NO it must be $\displaystyle 3x^2|y|\sqrt{10}$
You see that $\displaystyle y$ could be negative.

It make allot of sense , it's a variable therefore I don know the value.
So, I should always place variables in ||?
• Oct 14th 2011, 02:40 PM
Plato
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by vaironxxrd
So, I should always place variables in ||?

Well that depends.
$\displaystyle \sqrt{x^6y^4z^2}=|x^3|\cdot y^2\cdot|z|$.
Because odd powers can be negative but even powers are not negative.
• Oct 14th 2011, 02:42 PM
vaironxxrd
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by Plato
Just about all of that is incorrect.
$\displaystyle \sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$

Quote:

Originally Posted by e^(i*pi)
$\displaystyle 5 \times 4 \neq 25$

Indeed $\displaystyle 5 \times 5 = 25$ so you have $\displaystyle \sqrt{5^2x^2y}$

Quote:

Originally Posted by Plato
Well that depends.
$\displaystyle \sqrt{x^6y^4z^2}=|x^3|\cdot y^2\cdot|z|$.
Because odd powers can be negative but even powers are not negative.

Thanks a lot for the explanation, Plato.
Makes even more sense now!

I will watch a couple of videos doing radicals with ratios, and I might open a new thread. Thanks again
• Oct 18th 2011, 11:33 AM
hachataltoolimhakova
Re: Simplifying radical expressions with variables
Quote:

Originally Posted by HallsofIvy
The square root function is defined by "$\displaystyle \sqrt{a}$ is the non-negative number whose square is a" in order that it be single valued.

I have a dumb question. If I have the function $\displaystyle f(x,y) = \sqrt{25x^{2}y}$ I can have two solutions which yield non-negative numbers: $\displaystyle 5x\sqrt{y}$ for x>=0 and $\displaystyle -5x\sqrt{y}$ for x<=0. Are these valid solutions providing I state the ranges, or are they not valid because they do not satisfy all values of x? I assume the absolute value signs ensure that all values of x are valid. I was just wondering because by stating ranges for x, I can include -5 as a solution.
I have a dumb question. If I have the function $\displaystyle f(x,y) = \sqrt{25x^{2}y}$ I can have two solutions which yield non-negative numbers: $\displaystyle 5x\sqrt{y}$ for x>=0 and $\displaystyle -5x\sqrt{y}$ for x<=0. Are these valid solutions providing I state the ranges, or are they not valid because they do not satisfy all values of x? I assume the absolute value signs ensure that all values of x are valid. I was just wondering because by stating ranges for x, I can include -5 as a solution.
$\displaystyle f(x,y) = \sqrt{25x^{2}y}=5|x|\sqrt{y}$.
The domain is $\displaystyle x\in\mathbb{R}~\&~y\ge 0.$