layla asked in Science & MathematicsMathematics · 2 months ago

# 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

=>

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

• 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