SOLUTION: Solve x2 - 4x - k = 0. a) For what values of k the equation has only one solution? b) For what values of k are the solutions real? c) For what values of k are the solutions imag

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Solve x2 - 4x - k = 0. a) For what values of k the equation has only one solution? b) For what values of k are the solutions real? c) For what values of k are the solutions imag      Log On


   

Question 334936: Solve x2 - 4x - k = 0.
a) For what values of k the equation has only one solution?
b) For what values of k are the solutions real?
c) For what values of k are the solutions imaginary?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the discriminant.
Disc = b%5E2+-+4ac+=+16+%2B+4k
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If 16 + 4k = 0 --> 1 solution (or 2 equal solutions, x = 2, 2)
k = -4
--> x%5E2+-+4x+%2B+4+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-4x%2B4+=+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=%28-4%29%5E2-4%2A1%2A4=0.

Discriminant d=0 is zero! That means that there is only one solution: x+=+%28-%28-4%29%29%2F2%5C1.
Expression can be factored: 1x%5E2%2B-4x%2B4+=+%28x-2%29%2A%28x-2%29

Again, the answer is: 2, 2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B4+%29

x = 2, 2
===============
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If 16 + 4k < 0 --> 2 imaginary (complex conjugate) solutions
k < -4
eg k = -5
--> x%5E2+-+4x+%2B+5+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-4x%2B5+=+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=%28-4%29%5E2-4%2A1%2A5=-4.

The discriminant -4 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 -4 is + or - sqrt%28+4%29+=+2.

The solution is x%5B12%5D+=+%28--4%2B-i%2Asqrt%28+-4+%29%29%2F2%5C1+=++%28--4%2B-i%2A2%29%2F2%5C1+, or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B5+%29

------------
x = +2 ± i
=============
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If 16+4k > 0 --> 2 real solutions, different values
k > -4
eg k = -3
--> x%5E2+-+4x+%2B+3+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-4x%2B3+=+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=%28-4%29%5E2-4%2A1%2A3=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+4+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+4+%29%29%2F2%5C1+=+3
x%5B2%5D+=+%28-%28-4%29-sqrt%28+4+%29%29%2F2%5C1+=+1

Quadratic expression 1x%5E2%2B-4x%2B3 can be factored:
1x%5E2%2B-4x%2B3+=+%28x-3%29%2A%28x-1%29
Again, the answer is: 3, 1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B3+%29