# Use completion of squares to determine the root of the equation 3x^2 - 6x + 8 = 0

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### 3 Answers

Use the quadratic formula to find the solutions. In this case, the values are `a=3` , `b=-6` , and `c=8` .

`x=(-b+- sqrt(b^2-4ac))/(2a)`

` `where

`ax^2+bx+c=0`

Substitute in the values of `a=3` , `b=-6` , and `c=8` .

`x=(-(-6)+- sqrt((-6)^2-4(3)(8)))/(2(3))`

Multiply `-1` by each term inside the parentheses.

`x=(6+- sqrt((-6)^2-4(3)(8)))/(2(3))`

Simplify the section inside the radical.

`x=(6+-2i sqrt(15))/(2(3))`

Simplify the denominator of the quadratic formula.

`x=(6+-2i sqrt(15))/6`

Simplify the expression to solve for the `+` portion of the `+-.`

`x=(6+2i sqrt(15))/6`

Simplify the expression to solve for the` -` portion of the `+-.`

`x=(6-2i sqrt(15))/6`

The final answer is the combination of both solutions.

`x=(6+2i sqrt(15))/6, (6-2i sqrt(15))/6`

The quadratic equation 3x^2 - 6x + 8 = 0 has to be solved by the method completing the square.

The method relies on the identity `a^2 + 2a*b + b^2 = (a+b)^2`

For the given equation:

`3x^2 - 6x + 8 = 0`

Divide both the sides of the equation by 3.

`x^2 - 2x + 8/3 = 0`

=> `x^2 - 2*1*x + 1^2 = 1 - 8/3`

=> `(x - 1)^2 = -5/3`

`x - 1 = +-sqrt(-5/3)`

=> `x = 1 +- sqrt(-5/3)`

=> `x = 1 +- i*sqrt(5/3)`

**The solution of the equation `x^2 - 2x + 8/3 = 0` is **`x = 1 +- i*sqrt(5/3)`

The equation 3x^2 - 6x + 8 = 0 has to be solved for x.

3x^2 - 6x + 8 = 0

`(sqrt 3)^2*x^2 - sqrt 3*2*(3/sqrt 3)*x + 8 = 0`

`(sqrt 3)^2*x^2 - sqrt 3*2*(sqrt 3)*x + 8 = 0`

`(sqrt 3)^2*x^2 - sqrt 3*2*(sqrt 3)*x + (sqrt 3)^2 + 5 = 0`

`(sqrt 3*x - sqrt 3)^2 = -5`

`sqrt 3*x - sqrt 3 = +- sqrt 5*i`

`sqrt 3*x = sqrt 3 +- sqrt 5*i`

`x = (sqrt 3 +- sqrt 5*i)/sqrt 3`