SOLUTION: The braking distance (in feet) of a car going V mph is given by {{{d(v)=v^2/20+v}}} v is greater or equal to 0. how fast would the car have been traveling for a braking distance

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The braking distance (in feet) of a car going V mph is given by {{{d(v)=v^2/20+v}}} v is greater or equal to 0. how fast would the car have been traveling for a braking distance      Log On


   

Question 332228: The braking distance (in feet) of a car going V mph is given by d%28v%29=v%5E2%2F20%2Bv v is greater or equal to 0.
how fast would the car have been traveling for a braking distance of 150feet? round to nearest mile per hour.
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
d(v)=v^2/20+v
(v^2/20)+v=150
v^2+20v=3000
v^2+20v-3000=0
v=46 mph Quadratic formula below
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Ed
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B20x%2B-3000+=+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%2A1%2A-3000=12400.

Discriminant d=12400 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-20%2B-sqrt%28+12400+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2820%29%2Bsqrt%28+12400+%29%29%2F2%5C1+=+45.6776436283002
x%5B2%5D+=+%28-%2820%29-sqrt%28+12400+%29%29%2F2%5C1+=+-65.6776436283002

Quadratic expression 1x%5E2%2B20x%2B-3000 can be factored:
1x%5E2%2B20x%2B-3000+=+1%28x-45.6776436283002%29%2A%28x--65.6776436283002%29
Again, the answer is: 45.6776436283002, -65.6776436283002. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B20%2Ax%2B-3000+%29