SOLUTION: The height of an object thrown in the air is given by h = t2 – 7t + 3, where h is in feet and t is the time in seconds the object has been in motion. At what time (to the nearest

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The height of an object thrown in the air is given by h = t2 – 7t + 3, where h is in feet and t is the time in seconds the object has been in motion. At what time (to the nearest       Log On


   

Question 224402: The height of an object thrown in the air is given by h = t2 – 7t + 3, where h is in feet and t is the time in seconds the object has been in motion. At what time (to the nearest tenth) is the object 10 feet in the air?
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
h = t^2 – 7t + 3
Set h to zero and solve for t:
0 = t^2 – 7t + 3
Since we can't factor, we must resort to the quadratic equation. Doing so, yields:
x = {6.5, 0.5}
So, the object will be 10 feet at 0.5 seconds and 6.5 seconds.
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Details of quadratic equation follows...
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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%2B-7x%2B3+=+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=%28-7%29%5E2-4%2A1%2A3=37.

Discriminant d=37 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--7%2B-sqrt%28+37+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-7%29%2Bsqrt%28+37+%29%29%2F2%5C1+=+6.54138126514911
x%5B2%5D+=+%28-%28-7%29-sqrt%28+37+%29%29%2F2%5C1+=+0.45861873485089

Quadratic expression 1x%5E2%2B-7x%2B3 can be factored:
1x%5E2%2B-7x%2B3+=+1%28x-6.54138126514911%29%2A%28x-0.45861873485089%29
Again, the answer is: 6.54138126514911, 0.45861873485089. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-7%2Ax%2B3+%29