If (6√−2) is a root of the equation 3x^2−2kx+2=0, then find the value of k.?

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  • 2 months ago

    3x^2-2kx+2=0 has a root=6sqr(2)i

    =>

    3[6sqr(2)i]^2-2k[6sqr(2)i]+2=0

    =>

    -216-12sqr(2)ik+2=0

    =>

    k=-214/[12sqr(2)i]=107sqr(2)i/12 (answer)

    Thus,

    3x^2-107sqr(2)ix/6+2=0

    =>

    x=[107sqr(2)i]/6+/-sqr((107sqr(2)i/6)^2-24)]/6

    =>

    x=[107sqr(2)i/6+/-sqr(23762)i]/6

    =>

    x=sqr(2)i(107+/-109)/36

    =>

    x=6sqr(6)i or x=-sqr(2)i/18

     Check: [6sqr(2)i][-sqr(2)i/18]=12/18=2/3 valid.

  • 2 months ago

    3x^2−2kx+2=0

    3(6√(-2))^2 - 2k(6√(-2) + 2 = 0

    3 (36 * 2) - 12√(-2)k + 2 = 0

    216 - 12√(-2)k + 2 = 0

    218 - 12√(-2)k = 0

    - 12√(-2)k = - 218

      k = 109/i√(2) answer//

  • 2 months ago

    If (6√−2) is a root of the equation 3x^2 − 2kx + 2 = 0, 

    then find the value of k.

    3(6√−2)^2 - 2k (6√−2) + 2 = 0

    -214 - 12 i sqrt(2) k = 0

    Complex solution:

    k = (107 i)/(6 sqrt(2))

  • 2 months ago

    If (6√−2) is a root of the equation 3x^2−2kx+2=0  then --

    3 (6√−2)² - 2 k (6√−2) + 2  =  0

    => 3 ( - 72 ) -  (12√−2) k  + 2  =  0=> - 216 - (12√−2) k + 2 = 0

    => + (12√−2) k = - 214

    =>  k   =  - 214/(12√−2)  =  - (107/6√−2)  ......... Answer

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  • ted s
    Lv 7
    2 months ago

    Hmm....if x = 6√(-2) is a root then the coefficient of x has to be 0 ===> k = 0 ===> your equation is really 3 x² + 2 = 0 whose roots are NOT ± 6i √2...so correct your query and resubmit

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