SOLUTION: Please help me with this! 12x^2-5x-7, using completing the square. Thank you!

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Question 904911: Please help me with this!
12x^2-5x-7, using completing the square.
Thank you!

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert 12x%5E2%2B-5x%2B-7=0 to standard form by dividing both sides by 12:
We have: 1x%5E2%2B-0.416666666666667x%2B-0.583333333333333=0. What we want to do now is to change this equation to a complete square %28x%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28x%2Bsomenumber%29%5E2: %28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1x%5E2%2Bhighlight_red%28+-0.416666666666667%29+%2A+x%2B-0.583333333333333=0 that goes in front of x is -0.416666666666667, we know that -0.416666666666667=2*somenumber, or somenumber+=+-0.416666666666667%2F2. So, we know that our equation can be rewritten as %28x%2B-0.416666666666667%2F2%29%5E2+%2B+othernumber, and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that %28x%2B-0.416666666666667%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B-0.416666666666667x%2Bhighlight_green%28+-0.583333333333333+%29=0.


The highlighted red part must be equal to -0.583333333333333 (highlighted green part).

-0.416666666666667%5E2%2F4+%2B+othernumber+=+-0.583333333333333, or othernumber+=+-0.583333333333333--0.416666666666667%5E2%2F4+=+-0.626736111111111.
So, the equation converts to %28x%2B-0.416666666666667%2F2%29%5E2+%2B+-0.626736111111111+=+0, or %28x%2B-0.416666666666667%2F2%29%5E2+=+0.626736111111111.

Our equation converted to a square %28x%2B-0.416666666666667%2F2%29%5E2, equated to a number (0.626736111111111).

Since the right part 0.626736111111111 is greater than zero, there are two solutions:


, or





system%28+x+=+1%2C+x+=+-0.583333333333333+%29
Answer: x=1, -0.583333333333333.

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 12x%5E2%2B-5x%2B-7+=+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=%28-5%29%5E2-4%2A12%2A-7=361.

Discriminant d=361 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+361+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+361+%29%29%2F2%5C12+=+1
x%5B2%5D+=+%28-%28-5%29-sqrt%28+361+%29%29%2F2%5C12+=+-0.583333333333333

Quadratic expression 12x%5E2%2B-5x%2B-7 can be factored:
12x%5E2%2B-5x%2B-7+=+12%28x-1%29%2A%28x--0.583333333333333%29
Again, the answer is: 1, -0.583333333333333. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+12%2Ax%5E2%2B-5%2Ax%2B-7+%29