SOLUTION: {{{4k(k+5)=5}}}

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Question 439757: 4k%28k%2B5%29=5
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
4k(k+5)=5
4k^2+20k=5
4k^2+20k-5=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 4x%5E2%2B20x%2B-5+=+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=%2820%29%5E2-4%2A4%2A-5=480.

Discriminant d=480 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-20%2B-sqrt%28+480+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2820%29%2Bsqrt%28+480+%29%29%2F2%5C4+=+0.238612787525831
x%5B2%5D+=+%28-%2820%29-sqrt%28+480+%29%29%2F2%5C4+=+-5.23861278752583

Quadratic expression 4x%5E2%2B20x%2B-5 can be factored:
4x%5E2%2B20x%2B-5+=+4%28x-0.238612787525831%29%2A%28x--5.23861278752583%29
Again, the answer is: 0.238612787525831, -5.23861278752583. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B20%2Ax%2B-5+%29

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Ed