55) find the values of k if 4x^2 - kx + 9 is always positive.?

Answer 1

This is a minimum graph because the A(coefficient of x^2) is > 0. Thus, for it to be always positive, it must NOT have any roots(does not intersect y-axis at any point or above x-axis at all times). In other words, the discriminant(b^2-4ac) must be < 0.
b^2-4ac < 0
k^2-4(4)(9) < 0
k^2-144 < 0
(k-12)(k+12) < 0
-12 So, k can take values in between -12 and 12 i.e. -11,-10,...,10,11,12

Answer 2

4x^2 - kx + 9>0
(2x-3)^2 >0

y=(4ac-b^2/4a) >0
=(4*4*9-k^2)/(4*4)>0
4*4*9-k^2>0
4*4*9>k^2
2*2*3 >k
12>k

k={11,10,9,8,7,6,....}

Answer 3

for all positive value of k . bcoz it will be in form of (a-b)^2.
but k should be less than to ((2x)^2+(3)^2)/x