SOLUTION: Solve by completing the square: x^2 + 6x - 8 = 0 Solve by using the quadratic formula: z^2 -6z -14 = 0 Solve: x^2 + 2x - 8 <or=to 0

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Question 34247: Solve by completing the square:
x^2 + 6x - 8 = 0

Solve by using the quadratic formula:
z^2 -6z -14 = 0

Solve:
x^2 + 2x - 8
Answer by lyra(94) 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 1x%5E2%2B6x%2B-8=0 to standard form by dividing both sides by 1:
We have: 1x%5E2%2B6x%2B-8=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+6%29+%2A+x%2B-8=0 that goes in front of x is 6, we know that 6=2*somenumber, or somenumber+=+6%2F2. So, we know that our equation can be rewritten as %28x%2B6%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%2B6%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B6x%2Bhighlight_green%28+-8+%29=0.


The highlighted red part must be equal to -8 (highlighted green part).

6%5E2%2F4+%2B+othernumber+=+-8, or othernumber+=+-8-6%5E2%2F4+=+-17.
So, the equation converts to %28x%2B6%2F2%29%5E2+%2B+-17+=+0, or %28x%2B6%2F2%29%5E2+=+17.

Our equation converted to a square %28x%2B6%2F2%29%5E2, equated to a number (17).

Since the right part 17 is greater than zero, there are two solutions:

system%28+%28x%2B6%2F2%29+=+%2Bsqrt%28+17+%29%2C+%28x%2B6%2F2%29+=+-sqrt%28+17+%29+%29
, or

system%28+%28x%2B6%2F2%29+=+4.12310562561766%2C+%28x%2B6%2F2%29+=+-4.12310562561766+%29
system%28+x%2B6%2F2+=+4.12310562561766%2C+x%2B6%2F2+=+-4.12310562561766+%29
system%28+x+=+4.12310562561766-6%2F2%2C+x+=+-4.12310562561766-6%2F2+%29

system%28+x+=+1.12310562561766%2C+x+=+-7.12310562561766+%29
Answer: x=1.12310562561766, -7.12310562561766.


Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation az%5E2%2Bbz%2Bc=0 (in our case 1z%5E2%2B-6z%2B-14+=+0) has the following solutons:

z%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=%28-6%29%5E2-4%2A1%2A-14=92.

Discriminant d=92 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+92+%29%29%2F2%5Ca.

z%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+92+%29%29%2F2%5C1+=+7.79583152331272
z%5B2%5D+=+%28-%28-6%29-sqrt%28+92+%29%29%2F2%5C1+=+-1.79583152331272

Quadratic expression 1z%5E2%2B-6z%2B-14 can be factored:
1z%5E2%2B-6z%2B-14+=+1%28z-7.79583152331272%29%2A%28z--1.79583152331272%29
Again, the answer is: 7.79583152331272, -1.79583152331272. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B-14+%29


%28x%2B4%29%28x-2%29<=0
x%2B4=0
x=-4
x-2=0
x=2
so answer is (- infinity,2]
Hope this helps,
lyra