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Math Help - Simplifying radical expressions with variables

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    Senior Member vaironxxrd's Avatar
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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: \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?
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    Re: Simplifying radical expressions with variables

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

    Quote Originally Posted by Plato View Post
    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.
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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 \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 (\ -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; October 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 (\ -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.
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    Re: Simplifying radical expressions with variables

    Quote Originally Posted by Plato View Post
    Just about all of that is incorrect.
    \sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y
    Quote Originally Posted by e^(i*pi) View Post
    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}
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    Re: Simplifying radical expressions with variables

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

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

    Indeed 5 \times 5 = 25 so you have \sqrt{5^2x^2y}
    Quote Originally Posted by Plato View Post
    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 ||?
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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.
    \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.
    \sqrt{25x^2y}=\sqrt{5^2x^2y}=5|x|\sqrt y
    Quote Originally Posted by e^(i*pi) View Post
    5 \times 4 \neq 25

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

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