SOLUTION: Solve by completing the square 3x^2+5x-1=0

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Question 171916: Solve by completing the square
3x^2+5x-1=0
Found 2 solutions by jim_thompson5910, josmiceli:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

3x%5E2%2B5x-1 Start with the left side of the equation.


3%28x%5E2%2B%285%2F3%29x-1%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 5%2F3 to get 5%2F6. In other words, %281%2F2%29%285%2F3%29=5%2F6.


Now square 5%2F6 to get 25%2F36. In other words, %285%2F6%29%5E2=%285%2F6%29%285%2F6%29=25%2F36


3%28x%5E2%2B%285%2F3%29x%2Bhighlight%2825%2F36-25%2F36%29-1%2F3%29 Now add and subtract 25%2F36 inside the parenthesis. Make sure to place this after the "x" term. Notice how 25%2F36-25%2F36=0. So the expression is not changed.


3%28%28x%5E2%2B%285%2F3%29x%2B25%2F36%29-25%2F36-1%2F3%29 Group the first three terms.


3%28%28x%2B5%2F6%29%5E2-25%2F36-1%2F3%29 Factor x%5E2%2B%285%2F3%29x%2B25%2F36 to get %28x%2B5%2F6%29%5E2.


3%28%28x%2B5%2F6%29%5E2-37%2F36%29 Combine like terms.


3%28x%2B5%2F6%29%5E2%2B3%28-37%2F36%29 Distribute.


3%28x%2B5%2F6%29%5E2-37%2F12 Multiply.


So after completing the square, 3x%5E2%2B5x-1 transforms to 3%28x%2B5%2F6%29%5E2-37%2F12. So 3x%5E2%2B5x-1=3%28x%2B5%2F6%29%5E2-37%2F12.


So 3x%5E2%2B5x-1=0 is equivalent to 3%28x%2B5%2F6%29%5E2-37%2F12=0.


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3%28x%2B5%2F6%29%5E2-37%2F12=0 Start with the given equation.


3%28x%2B5%2F6%29%5E2=0%2B37%2F12Add 37%2F12 to both sides.


3%28x%2B5%2F6%29%5E2=37%2F12 Combine like terms.


%28x%2B5%2F6%29%5E2=%2837%2F12%29%2F%283%29 Divide both sides by 3.


%28x%2B5%2F6%29%5E2=37%2F36 Reduce.


x%2B5%2F6=0%2B-sqrt%2837%2F36%29 Take the square root of both sides.


x%2B5%2F6=sqrt%2837%2F36%29 or x%2B5%2F6=-sqrt%2837%2F36%29 Break up the "plus/minus" to form two equations.


x%2B5%2F6=sqrt%2837%29%2F6 or x%2B5%2F6=-sqrt%2837%29%2F6 Simplify the square root.


x=-5%2F6%2Bsqrt%2837%29%2F6 or x=-5%2F6-sqrt%2837%29%2F6 Subtract 5%2F6 from both sides.


x=%28-5%2Bsqrt%2837%29%29%2F%286%29 or x=%28-5-sqrt%2837%29%29%2F%286%29 Combine the fractions.


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Answer:


So the solutions are x=%28-5%2Bsqrt%2837%29%29%2F%286%29 or x=%28-5-sqrt%2837%29%29%2F%286%29.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
First add 1 to both sides
3x%5E2+%2B+5x+=+1
Now divide both sides by 3
x%5E2+%2B+%285%2F3%29%2Ax+=+1%2F3
Take 1/2 of the coefficient of x, square it
and add it to both sides
x%5E2+%2B+%285%2F3%29%2Ax+%2B+%285%2F6%29%5E2+=+%285%2F6%29%5E2+%2B+1%2F3
The thing to realize at this point is that the
left side is a perfect square. It is
%28x+%2B+5%2F6%29%5E2, so
%28x+%2B+5%2F6%29%5E2+=+25%2F36+%2B+12%2F36
%28x+%2B+5%2F6%29%5E2+=+37%2F36
Take the square root of both sides
x+%2B+5%2F6+=+sqrt%2837%29%2F6
Solve for x
x+=+%28sqrt%2837%29+-+5%29+%2F+6
and
x+=+%28-sqrt%2837%29+-+5%29+%2F+6
Both answers are correct