Question

if 3x^2+4kx+1>0 for all real values of x then k lies in the interval​

Answer 1

Step-by-step explanation:

3x² + 4kx + 1 > 0 for all real value of x

this is possible only when,

D = b² -4ac< 0

( 4 k)² - 4(3) < 0

16K² -12 < 0

4K² -3 < 0

(2K-√3)(2K+√3) < 0

hope it's help u ✌️✌️✌️

-√3/2 < K < √3/2

Answer 2

Step-by-step explanation:

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