If 3x^2+4kx +1 >0 for all real values of x , then k lies in the interval
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Answered by
22
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
-√3/2 < K < √3/2
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
-√3/2 < K < √3/2
Ramu111:
why b^2-4ac is <0
why b^2-4ac <0
why b^2-4ac <0 please say fast
ax² + bx+ c will be negative only when, a< 0 and D< 0
ax² + bx + c will be positive only when , a> 0 and D< 0
you best understand of this if you see this graph
because ax² + bx + c will be positive means this will lies above from x -axis
so, no real roots available for this
it means D< 0
thank you so much
Answered by
9
If the discriminant is negative then the roots of the quadratic function are imaginary.
So the quadratic function is never 0 for any real x.
Then as the coefficient of x^2 is positive , the value of the quadratic is positive for some large x for sure. So it will be positive for all x.
Discriminant = 16 k^2 - 12 < 0
So. The range of k is
- sqrt(3)/2 < k < sqrt(3)/2
Answer.
So the quadratic function is never 0 for any real x.
Then as the coefficient of x^2 is positive , the value of the quadratic is positive for some large x for sure. So it will be positive for all x.
Discriminant = 16 k^2 - 12 < 0
So. The range of k is
- sqrt(3)/2 < k < sqrt(3)/2
Answer.
Click on the red heart thanks above please
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