SOLUTION: state all the values of k that would make the following have imaginary roots: x^2-9x+k=0 please show work

Question 253227: state all the values of k that would make the following have imaginary roots: x^2-9x+k=0
please show work
Found 2 solutions by drk, richwmiller:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
We are given the original equation as:
(i)
--
First, we should solve for x using the quadratic formula:
(ii)
replacing a, b, and c with 1, -9, and k, we get
(iii)
Now we want imaginary answers, which means the discriminant must < 0. So we get

Solving for k, we get

check:
Let k = 21.
From (i) we get

from (ii), we get

and we can see that we get a negative square root. This tells us that our answer is correct.

Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 5, 4. Here's your graph:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=0 is zero! That means that there is only one solution: .
Expression can be factored:

Again, the answer is: 4.5, 4.5. Here's your graph:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -1 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

In the field of imaginary numbers, the square root of -1 is + or - .

The solution is

Here's your graph:

where b^2-4ak<0
a=1
b=-9
81-4(1)k=0
81-4k=0
81=4k
81/4=k
if k> 81/4 then there will be no real solutions

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