SOLUTION: 4x^2-4x+3=0 Move the constant to the right side of the equation. Multiply each term in the equation by four times the coefficient of the x^2 term. Square the coeffici

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: 4x^2-4x+3=0 Move the constant to the right side of the equation. Multiply each term in the equation by four times the coefficient of the x^2 term. Square the coeffici      Log On


   

Question 252024: 4x^2-4x+3=0
Move the constant to the right side of the equation.
Multiply each term in the equation by four times the coefficient of the x^2 term.
Square the coefficient of the original x term and add it to both sides of the equation
Take the square root of both sides.
Set the left side of the equation equal to the positive square root of the number on the right side and solve for x
Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
I am assuming the steps you mention are what you have tried.
The second should be divide both sides by 4 (the coefficient of x^2)
third step should be take 1/2 the coefficient of x and square it and then add it to both sides
Possible intermediate steps:
4 x^2-4 x+3 = 0
Solve the quadratic equation by completing the square:
Divide both sides by 4:
x^2-x+3/4 = 0
Subtract 3/4 from both sides:
x^2-x = -3/4
Add 1/4 to both sides:
x^2-x+1/4 = -1/2
Factor the left hand side:
(x-1/2)^2 = -1/2
Take the square root of both sides:
|x-1/2| = i/sqrt(2)
Eliminate the absolute value:
x-1/2 = -i/sqrt(2) or x-1/2 = i/sqrt(2)
Add 1/2 to both sides:
x = 1/2 (1-i sqrt(2)) or x-1/2 = i/sqrt(2)
Add 1/2 to both sides:
x = 1/2 (1-i sqrt(2)) or x = 1/2 (1+i sqrt(2))
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 4x%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%2A4%2A3=-32.

The discriminant -32 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 -32 is + or - sqrt%28+32%29+=+5.65685424949238.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B-4%2Ax%2B3+%29

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B0.75+=+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=%28-1%29%5E2-4%2A1%2A0.75=-2.

The discriminant -2 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 -2 is + or - sqrt%28+2%29+=+1.4142135623731.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B0.75+%29

4x^2-4x+3=0