SOLUTION: Please help me solve this equation 5-5 Completing the Square Solving a quadratic equation if the coefficient of x^2 is NOT 1 3x^2+12x-9=0

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Question 191084: Please help me solve this equation
5-5 Completing the Square
Solving a quadratic equation if the coefficient of x^2 is NOT 1

3x^2+12x-9=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First, let's complete the square for the left side of the equation 3x%5E2%2B12x-9=0



3x%5E2%2B12x-9 Start with the given expression.


3%28x%5E2%2B4x-3%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 4 to get 2. In other words, %281%2F2%29%284%29=2.


Now square 2 to get 4. In other words, %282%29%5E2=%282%29%282%29=4


3%28x%5E2%2B4x%2Bhighlight%284-4%29-3%29 Now add and subtract 4 inside the parenthesis. Make sure to place this after the "x" term. Notice how 4-4=0. So the expression is not changed.


3%28%28x%5E2%2B4x%2B4%29-4-3%29 Group the first three terms.


3%28%28x%2B2%29%5E2-4-3%29 Factor x%5E2%2B4x%2B4 to get %28x%2B2%29%5E2.


3%28%28x%2B2%29%5E2-7%29 Combine like terms.


3%28x%2B2%29%5E2%2B3%28-7%29 Distribute.


3%28x%2B2%29%5E2-21 Multiply.


So after completing the square, 3x%5E2%2B12x-9 transforms to 3%28x%2B2%29%5E2-21. So 3x%5E2%2B12x-9=3%28x%2B2%29%5E2-21.


So 3x%5E2%2B12x-9=0 is equivalent to 3%28x%2B2%29%5E2-21=0.


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Now let's solve 3%28x%2B2%29%5E2-21=0


3%28x%2B2%29%5E2-21=0 Start with the given equation.


3%28x%2B2%29%5E2=21 Add 21 to both sides


%28x%2B2%29%5E2=%2821%29%2F%283%29 Divide both sides by 3.


%28x%2B2%29%5E2=7 Reduce.


x%2B2=%22%22%2B-sqrt%287%29 Take the square root of both sides. (note: don't forget the "plus/minus")


x%2B2=sqrt%287%29 or x%2B2=-sqrt%287%29 Break up the "plus/minus" to form two equations.


x=-2%2Bsqrt%287%29 or x=-2-sqrt%287%29 Subtract 2 from both sides.


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Answer:


So the solutions are x=-2%2Bsqrt%287%29 or x=-2-sqrt%287%29.