# Thread: Simplifying radical expressions with variables

1. ## Simplifying radical expressions with variables

Hello forum vaironxxrd here.

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

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

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

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

2. ## Re: Simplifying radical expressions with variables

Originally Posted by vaironxxrd
Problem: $\sqrt{25x^2y}$
Solution: $\sqrt{25x^2y}$ = $x\sqrt{5 \cdot 4 y}$
= $2x\sqrt{5y}$
Is that solution right? or what am I doing wrong?
Just about all of that is incorrect.
$\sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$

3. ## Re: Simplifying radical expressions with variables

Originally Posted by vaironxxrd
Hello forum vaironxxrd here.

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

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

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

Is that solution right? or what am I doing wrong?
$5 \times 4 \neq 25$

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

4. ## Re: Simplifying radical expressions with variables

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

5. ## Re: Simplifying radical expressions with variables

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 $\left( {\forall x} \right)\left( {\forall y \geqslant 0} \right)\left[ {\sqrt {25x^2 y} \geqslant 0} \right]$

6. ## 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 $(\ -5 \left \mid x \mid \left \sqrt{y})^2$ 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?

7. ## Re: Simplifying radical expressions with variables

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 $(\ -5 \left \mid x \mid \left \sqrt{y})^2$ 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 $2^2= 4$ and $(-2)^2= 4$. The square root function is defined by " $\sqrt{a}$ is the non-negative number whose square is a" in order that it be single valued.

8. ## Re: Simplifying radical expressions with variables

Originally Posted by Plato
Just about all of that is incorrect.
$\sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$
Originally Posted by e^(i*pi)
$5 \times 4 \neq 25$

Indeed $5 \times 5 = 25$ so you have $\sqrt{5^2x^2y}$
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: $\sqrt{90x^4y^2}$

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

$3x^2y\sqrt{5 \cdot 2}$ = $3x^2y\sqrt{10}$

9. ## Re: Simplifying radical expressions with variables

Originally Posted by vaironxxrd
Problem: $\sqrt{90x^4y^2}$
Solution: $3x^2{\color{red}y}\sqrt{10}$
NO it must be $3x^2|y|\sqrt{10}$
You see that $y$ could be negative.

10. ## Re: Simplifying radical expressions with variables

Originally Posted by Plato
Just about all of that is incorrect.
$\sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$
Originally Posted by e^(i*pi)
$5 \times 4 \neq 25$

Indeed $5 \times 5 = 25$ so you have $\sqrt{5^2x^2y}$
Originally Posted by Plato
NO it must be $3x^2|y|\sqrt{10}$
You see that $y$ 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 ||?

11. ## Re: Simplifying radical expressions with variables

Originally Posted by vaironxxrd
So, I should always place variables in ||?
Well that depends.
$\sqrt{x^6y^4z^2}=|x^3|\cdot y^2\cdot|z|$.
Because odd powers can be negative but even powers are not negative.

12. ## Re: Simplifying radical expressions with variables

Originally Posted by Plato
Just about all of that is incorrect.
$\sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$
Originally Posted by e^(i*pi)
$5 \times 4 \neq 25$

Indeed $5 \times 5 = 25$ so you have $\sqrt{5^2x^2y}$
Originally Posted by Plato
Well that depends.
$\sqrt{x^6y^4z^2}=|x^3|\cdot y^2\cdot|z|$.
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

13. ## Re: Simplifying radical expressions with variables

Originally Posted by HallsofIvy
The square root function is defined by " $\sqrt{a}$ 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 $f(x,y) = \sqrt{25x^{2}y}$ I can have two solutions which yield non-negative numbers: $5x\sqrt{y}$ for x>=0 and $-5x\sqrt{y}$ 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 $f(x,y) = \sqrt{25x^{2}y}$ I can have two solutions which yield non-negative numbers: $5x\sqrt{y}$ for x>=0 and $-5x\sqrt{y}$ 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.
$f(x,y) = \sqrt{25x^{2}y}=5|x|\sqrt{y}$.
The domain is $x\in\mathbb{R}~\&~y\ge 0.$