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Thread: Simplifying radical expressions with variables

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    Simplifying radical expressions with variables

    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?
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by vaironxxrd View Post
    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.
    $\displaystyle \sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y$
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by vaironxxrd View Post
    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 $\displaystyle \sqrt{5^2x^2y}$
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by Plato View Post
    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.
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by hachataltoolimhakova View Post
    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 $\displaystyle \left( {\forall x} \right)\left( {\forall y \geqslant 0} \right)\left[ {\sqrt {25x^2 y} \geqslant 0} \right]$
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    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?
    Last edited by hachataltoolimhakova; Oct 14th 2011 at 10:11 AM.
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by hachataltoolimhakova View Post
    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.
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by Plato View Post
    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) View Post
    $\displaystyle 5 \times 4 \neq 25$

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


    $\displaystyle 3x^2y\sqrt{5 \cdot 2}$ = $\displaystyle 3x^2y\sqrt{10}$
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by vaironxxrd View Post
    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.
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by Plato View Post
    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) View Post
    $\displaystyle 5 \times 4 \neq 25$

    Indeed $\displaystyle 5 \times 5 = 25$ so you have $\displaystyle \sqrt{5^2x^2y}$
    Quote Originally Posted by Plato View Post
    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 ||?
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by vaironxxrd View Post
    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.
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by Plato View Post
    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) View Post
    $\displaystyle 5 \times 4 \neq 25$

    Indeed $\displaystyle 5 \times 5 = 25$ so you have $\displaystyle \sqrt{5^2x^2y}$
    Quote Originally Posted by Plato View Post
    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
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by HallsofIvy View Post
    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.
    Thanks in advance.
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by hachataltoolimhakova View Post
    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.
    You do not need cases. Just write:
    $\displaystyle f(x,y) = \sqrt{25x^{2}y}=5|x|\sqrt{y}$.
    The domain is $\displaystyle x\in\mathbb{R}~\&~y\ge 0.$
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