SOLUTION: Solve the quadratic equation by completing the square: 3x^2+6x-4=0

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Question 169929: Solve the quadratic equation by completing the square:
3x^2+6x-4=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

3x%5E2%2B6x-4 Start with the right side of the equation.


3%28x%5E2%2B2x-4%2F3%29 Factor out the x%5E2 coefficient 3. This step is very important: the x%5E2 coefficient must be equal to 1.


Take half of the x coefficient 2 to get 1. In other words, %281%2F2%29%282%29=1.


Now square 1 to get 1. In other words, %281%29%5E2=%281%29%281%29=1


3%28x%5E2%2B2x%2Bhighlight%281-1%29-4%2F3%29 Now add and subtract 1 inside the parenthesis. Make sure to place this after the "x" term. Notice how 1-1=0. So the expression is not changed.


3%28%28x%5E2%2B2x%2B1%29-1-4%2F3%29 Group the first three terms.


3%28%28x%2B1%29%5E2-1-4%2F3%29 Factor x%5E2%2B2x%2B1 to get %28x%2B1%29%5E2.


3%28%28x%2B1%29%5E2-7%2F3%29 Combine like terms.


3%28x%2B1%29%5E2%2B3%28-7%2F3%29 Distribute.


3%28x%2B1%29%5E2-7 Multiply.


So after completing the square, 3x%5E2%2B6x-4 transforms to 3%28x%2B1%29%5E2-7. So 3x%5E2%2B6x-4=3%28x%2B1%29%5E2-7.


So 3x%5E2%2B6x-4=0 is equivalent to 3%28x%2B1%29%5E2-7=0.


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3%28x%2B1%29%5E2-7=0 Start with the given equation.


3%28x%2B1%29%5E2=0%2B7Add 7 to both sides.


3%28x%2B1%29%5E2=7 Combine like terms.


%28x%2B1%29%5E2=%287%29%2F%283%29 Divide both sides by 3.


x%2B1=0%2B-sqrt%287%2F3%29 Take the square root of both sides.


x%2B1=sqrt%287%2F3%29 or x%2B1=-sqrt%287%2F3%29 Break up the "plus/minus" to form two equations.


x%2B1=sqrt%2821%29%2F3 or x%2B1=-sqrt%2821%29%2F3 Simplify the square root.


x=-1%2Bsqrt%2821%29%2F3 or x=-1-sqrt%2821%29%2F3 Subtract 1 from both sides.


x=%28-3%2Bsqrt%2821%29%29%2F%283%29 or x=%28-3-sqrt%2821%29%29%2F%283%29 Combine the fractions.


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Answer:


So the solutions are x=%28-3%2Bsqrt%2821%29%29%2F%283%29 or x=%28-3-sqrt%2821%29%29%2F%283%29.