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...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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.**