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

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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) About Me  (Show Source):
You can put this solution on YOUR website!
We are given the original equation as:
(i) x%5E2-9x%2Bk=0
--
First, we should solve for x using the quadratic formula:
(ii) x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
replacing a, b, and c with 1, -9, and k, we get
(iii) x+=+%289+%2B-+sqrt%28+81-4%2Ak+%29%29%2F%282%29+
Now we want imaginary answers, which means the discriminant must < 0. So we get
81+-+4k+%3C+0
Solving for k, we get
k+%3E+20.25
check:
Let k = 21.
From (i) we get
x%5E2+-9x+%2B+21+=+0
from (ii), we get
x+=+%289+%2B-+sqrt%28+81-84+%29%29%2F%282%29+
and we can see that we get a negative square root. This tells us that our answer is correct.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-9x%2B20+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-9%29%5E2-4%2A1%2A20=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--9%2B-sqrt%28+1+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-9%29%2Bsqrt%28+1+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%28-9%29-sqrt%28+1+%29%29%2F2%5C1+=+4

Quadratic expression 1x%5E2%2B-9x%2B20 can be factored:
1x%5E2%2B-9x%2B20+=+1%28x-5%29%2A%28x-4%29
Again, the answer is: 5, 4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-9%2Ax%2B20+%29

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-9x%2B20.25+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-9%29%5E2-4%2A1%2A20.25=0.

Discriminant d=0 is zero! That means that there is only one solution: x+=+%28-%28-9%29%29%2F2%5C1.
Expression can be factored: 1x%5E2%2B-9x%2B20.25+=+1%28x-4.5%29%2A%28x-4.5%29

Again, the answer is: 4.5, 4.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-9%2Ax%2B20.25+%29

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-9x%2B20.5+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-9%29%5E2-4%2A1%2A20.5=-1.

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 - sqrt%28+1%29+=+1.

The solution is x%5B12%5D+=+%28--9%2B-+i%2Asqrt%28+-1+%29%29%2F2%5C1+=++%28--9%2B-+i%2A1%29%2F2%5C1+

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-9%2Ax%2B20.5+%29

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