SOLUTION: How do you do a quadratic equation not ending in zero? The problem on my worksheet is 3xsquared + 3x = 4 ? I dont know what to do.

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Question 293599: How do you do a quadratic equation not ending in zero? The problem on my worksheet is 3xsquared + 3x = 4 ? I dont know what to do.
Found 2 solutions by jim_thompson5910, richwmiller:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2%2B3x=4 Start with the given equation.


3x%5E2%2B3x-4=0 Subtract 4 from both sides.


Now you just need to solve 3x%5E2%2B3x-4=0. Does that help?

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
3x^2+3x=4
One approach is to make it equal to zero.
3x^2+3x-4=0
Another is complete the square
add 4
divide by 3
x^2+x=4/3
add (1/2)^2=1/4
x^2+x+1/4=4/3+1/4
(x+1/2)^2=16/12+3/12
(x+1/2)^2=19/12
I leave the rest for you
We could also use the quadratic formula.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B3x%2B-4+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%283%29%5E2-4%2A3%2A-4=57.

Discriminant d=57 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+57+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+57+%29%29%2F2%5C3+=+0.758305739211792
x%5B2%5D+=+%28-%283%29-sqrt%28+57+%29%29%2F2%5C3+=+-1.75830573921179

Quadratic expression 3x%5E2%2B3x%2B-4 can be factored:
3x%5E2%2B3x%2B-4+=+3%28x-0.758305739211792%29%2A%28x--1.75830573921179%29
Again, the answer is: 0.758305739211792, -1.75830573921179. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B3%2Ax%2B-4+%29