SOLUTION: Consider the function f(x) = e^rx. Determine the values of r so that f sasisfies the equation f"(x) + f'(x)−6f(x) = 0. [

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Question 1049702: Consider the function f(x) = e^rx. Determine the values of r so that f sasisfies the equation f"(x) + f'(x)−6f(x) = 0.
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Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
Consider the function f%28x%29 = e%5Erx. Determine the values of r so that f satisfies the equation f"(x) + f'(x)-6f(x) = 0
solution
f'(x)=r%2Ae%5Erx
f"(x)=r%5E2%2Ae%5Erx

f"(x) + f'(x)-6f(x) = 0
r%5E2%2Ae%5Erx+r%2Ae%5Erx-6e%5Erx=0
e%5Erx*%28r%5E2%2Br-6%29=0
e%5Erx=0 and r%5E2%2Br-6=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ar%5E2%2Bbr%2Bc=0 (in our case 1r%5E2%2B1r%2B-6+=+0) has the following solutons:

r%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=%281%29%5E2-4%2A1%2A-6=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+25+%29%29%2F2%5Ca.

r%5B1%5D+=+%28-%281%29%2Bsqrt%28+25+%29%29%2F2%5C1+=+2
r%5B2%5D+=+%28-%281%29-sqrt%28+25+%29%29%2F2%5C1+=+-3

Quadratic expression 1r%5E2%2B1r%2B-6 can be factored:
1r%5E2%2B1r%2B-6+=+1%28r-2%29%2A%28r--3%29
Again, the answer is: 2, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-6+%29


so
r=-3
r=2