Results 1 to 5 of 5

Math Help - Not sure what this comes under.

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    149

    Not sure what this comes under.

    Had this question in a test and got it wrong.

    find the solutions to  2\sqrt{x^5}-72\sqrt{x}=0

    answer is x = 0 , x=6 , or x = -6; not sure what i do with this equation , im on a study break at moment so im trying to get on top of these things before i get back. You help is appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,963
    Thanks
    1631
    Quote Originally Posted by el123 View Post
    Had this question in a test and got it wrong.

    find the solutions to  2\sqrt{x^5}-72\sqrt{x}=0

    answer is x = 0 , x=6 , or x = -6; not sure what i do with this equation , im on a study break at moment so im trying to get on top of these things before i get back. You help is appreciated
    I don't know where you got "x= 0, x= 6, or x=-6" but 6 and -6 are obviously NOT solutions. If x= 6, for example, then x^5= 6^5= 7776 so \sqrt{x^5}= \sqrt{7776}= 88.816 while [tex]\sqrt{6}= 2.44945 so 2\sqrt{x^5}- 72\sqrt{x} = 2(88.816)- 72(2.44945)= 1.2716, not 0. Did you mean "x= 0, x^2= 6, or x^2= -6"?

    \sqrt{x^5}= \left(\sqrt{x}\right)^5 so let u= \sqrt{x} and the equation becomes 2u^5- 72u= 0. That can be factored as 2u(u^4- 36)= 2u(u^2- 6)(u^2+ 6)= 0. Therefore, we have u= 0 or u^2- 6= 0 or u^2+ 6= 0. The 5 solutions to this fifth degree equation are x= 0, x= \sqrt{6}, x= -\sqrt{6}, x= i\sqrt{6}, and x= -i\sqrt{6}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2009
    Posts
    149
    Were not using imaginary numbers in this course , and the answer is the one i gave because we have been given the answers for it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288
    Actually, x=0,-6,6 are solutions.

    How?

    2\sqrt{x^5}=72\sqrt{x}
    \sqrt{x^5}=36\sqrt{x}
    x^5=1296x
    x^4=1296
    x=6,-6
    And x=0 is an obvious solution.


    Put into the equation
    2\sqrt{x^5}=176.363...
    72\sqrt{6}=176.363...
    Hence:  2\sqrt{6^5}-72\sqrt{6}=0

    Actually, your values of x=\sqrt{6} are wrong
    2\sqrt{\sqrt{6}}=3.130...
    72\sqrt{\sqrt{6}}=112.6860...
    And so on with your other values

    HallsofIvy, here's your mistake
    2u(u^2-6)(u^2+6)=0
    u^2-6=0
    u=\sqrt{6}
    u=\sqrt{x}
    Hence
    \sqrt{x}=\sqrt{6}
    x=6
    And so on
    Last edited by I-Think; September 1st 2009 at 05:07 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,867
    Thanks
    640
    Quote Originally Posted by I-Think View Post
    Actually, x=0,-6,6 are solutions.
    x = -6 is not a solution ... for real solutions, the domain of the equation is x \ge 0
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum