SOLUTION: What three techniques can be used to solve a quadratic equation, demonstrate each using x^2-10x-39=0

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Question 203639: What three techniques can be used to solve a quadratic equation, demonstrate each using x^2-10x-39=0
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
To solve:
x^2-10x-39=0
.
Three techniques could be:
1. factoring
2. completing the square
3. quadratic equation
.
Three techniques could be:
1. factoring
x^2-10x-39=0
(x-13)(x+3)=0
x = {-3, 13}
.
2. completing the square
x^2-10x-39=0
x^2-10x = 39
x^2-10x+25 = 39+25
(x-5)^2 = 64
(x-5) = sqrt(64)
x-5 = +-(8)
x = +-(8)+5
x = {-3, 13}
.
3. quadratic equation
x^2-10x-39=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-10x%2B-39+=+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=%28-10%29%5E2-4%2A1%2A-39=256.

Discriminant d=256 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--10%2B-sqrt%28+256+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-10%29%2Bsqrt%28+256+%29%29%2F2%5C1+=+13
x%5B2%5D+=+%28-%28-10%29-sqrt%28+256+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B-10x%2B-39 can be factored:
1x%5E2%2B-10x%2B-39+=+1%28x-13%29%2A%28x--3%29
Again, the answer is: 13, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-10%2Ax%2B-39+%29
.