find the value of k in each of the following?a.) kx^2-6x+1=0 if the equation has real, rational, and equal roots.

a.) kx^2-6x+1=0 if the equation has real, rational, and equal
roots.

If the function has rational , real and equal roots, then
that mean delta will be 0:

We know
that:

delta = b^2 - 4ac

==>
delta = (-6)^2 - 4*k*1

==> 36 - 4k =
0

==> -4k = -36

==> k =
-36/-4 = 9

==> k = 9 

b.) 2x^2+6x+k=0 find the smallest integral value of
k, such that the equation had imaginary roots

If the function has
imaginary of (complex) roots, then delta should be less than 0 (
negative).

==> delta = b^2 -
4ac

==> delta = (6^2) - 4*2*k <
0

==> 36 - 8k <
0

==> -8k <
-36

==> k >
-36/-8

==> k > 9/2

Then
the smallest integer is the first integer that comes after 9/2

9/2 =
4.5 , then the next integer s 5

Then k = 5 is the
smallest integer so the equation has complex roots.