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

### 5 Answers

- PinkgreenLv 72 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.

- Engr. RonaldLv 72 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//

- KrishnamurthyLv 72 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))

- Pramod KumarLv 72 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 sLv 72 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