SOLUTION: Solve using completing the square? -3x^2+6x-5

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Question 207639: Solve using completing the square? -3x^2+6x-5
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

-3x%5E2%2B6x-5 Start with the given expression.


-3%28x%5E2-2x%2B5%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%28-2%29=-1.


Now square -1 to get 1. In other words, %28-1%29%5E2=%28-1%29%28-1%29=1


-3%28x%5E2-2x%2Bhighlight%281-1%29%2B5%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%5E2-2x%2B1%29-1%2B5%2F3%29 Group the first three terms.


-3%28%28x-1%29%5E2-1%2B5%2F3%29 Factor x%5E2-2x%2B1 to get %28x-1%29%5E2.


-3%28%28x-1%29%5E2%2B2%2F3%29 Combine like terms.


-3%28x-1%29%5E2-3%282%2F3%29 Distribute.


-3%28x-1%29%5E2-2 Multiply.


So after completing the square, -3x%5E2%2B6x-5 transforms to -3%28x-1%29%5E2-2. So -3x%5E2%2B6x-5=-3%28x-1%29%5E2-2.


So -3x%5E2%2B6x-5=0 is equivalent to -3%28x-1%29%5E2-2=0.


So we can solve -3%28x-1%29%5E2-2=0 in place of solving -3x%5E2%2B6x-5=0


-3%28x-1%29%5E2-2=0 Start with the given equation.


-3%28x-1%29%5E2=0%2B2Add 2 to both sides.


-3%28x-1%29%5E2=2 Combine like terms.


%28x-1%29%5E2=%282%29%2F%28-3%29 Divide both sides by -3.


%28x-1%29%5E2=-2%2F3 Reduce.


x-1=%22%22%2B-sqrt%28-2%2F3%29 Take the square root of both sides.


x-1=sqrt%28-2%2F3%29 or x-1=-sqrt%28-2%2F3%29 Break up the "plus/minus" to form two equations.


x-1=%28i%2Asqrt%286%29%29%2F3 or x-1=-%28i%2Asqrt%286%29%29%2F3 Simplify the square root. Note: i=sqrt%28-1%29


x=1%2B%28i%2Asqrt%286%29%29%2F3 or x=1-%28i%2Asqrt%286%29%29%2F3 Add 1 to both sides.


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Answer:


So the solutions are x=1%2B%28i%2Asqrt%286%29%29%2F3 or x=1-%28i%2Asqrt%286%29%29%2F3.