SOLUTION: i. f(x)=x^2-4x+4 ii. f(x)=3x^2-4x^2+2

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Question 632274: i. f(x)=x^2-4x+4
ii. f(x)=3x^2-4x^2+2
Answer by reviewermath(1029) 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-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+=+1%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

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B-4x%2B2+=+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%2A3%2A2=-8.

The discriminant -8 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 -8 is + or - sqrt%28+8%29+=+2.82842712474619.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B-4%2Ax%2B2+%29