SOLUTION: After t sec, the height of a rocket launched straight upward from the ground level with an initial velocity of 76ft/sec can be modeled by the polynomial 76t-16t^2. After how many s

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: After t sec, the height of a rocket launched straight upward from the ground level with an initial velocity of 76ft/sec can be modeled by the polynomial 76t-16t^2. After how many s      Log On


   

Question 899127: After t sec, the height of a rocket launched straight upward from the ground level with an initial velocity of 76ft/sec can be modeled by the polynomial 76t-16t^2. After how many seconds will the rocket reach a height of 18 feet above the launch (that is, equal to +18)?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
-16t^2 + 76t = 18
-16t^2 + 76t - 18 = 0
x = 0.25, 4.5
.25sec after launch, rocket reaches 18ft
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B76x%2B-18+=+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=%2876%29%5E2-4%2A-16%2A-18=4624.

Discriminant d=4624 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-76%2B-sqrt%28+4624+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2876%29%2Bsqrt%28+4624+%29%29%2F2%5C-16+=+0.25
x%5B2%5D+=+%28-%2876%29-sqrt%28+4624+%29%29%2F2%5C-16+=+4.5

Quadratic expression -16x%5E2%2B76x%2B-18 can be factored:
-16x%5E2%2B76x%2B-18+=+-16%28x-0.25%29%2A%28x-4.5%29
Again, the answer is: 0.25, 4.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B76%2Ax%2B-18+%29