SOLUTION: Find the perfect square trinomial whose first two terms are: a^2 - 8a Solve by completing the square. t^2 - 3t - 7 = 0 An object in launched directly upward at 64 feet per

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Question 201101: Find the perfect square trinomial whose first two terms are:
a^2 - 8a
Solve by completing the square.
t^2 - 3t - 7 = 0
An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet
high. What will be the object's maximum height? When will it attain this height?

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
1) To find the perfect square trinomial, add the square of half the x-coefficient .
a%5E2-8a%2B%288%2F2%29%5E2 Simplify.
highlight%28a%5E2-8a%2B16+=+%28a-4%29%5E2%29
2) Solve by completing the square:
t%5E2-3t-7+=+0 Add 7 to both sides.
t%5E2-3t+=+7 Now complete the square in t by adding the square of half the t-coefficient to both sides. This is %283%2F2%29%5E2+=+9%2F4.
t%5E2-3t%2B9%2F4+=+7%2B%289%2F4%29 Factor the left side and simplify the right side.
%28t-%283%2F2%29%29%5E2+=+%2828%2F4%29%2B%289%2F4%29
%28t-%283%2F2%29%29%5E2+=+37%2F4 Take the square root of both sides.
t-%283%2F2%29+=+%28sqrt%2837%29%29%2F2 or t-%283%2F2%29+=+-%28sqrt%2837%29%29%2F2 Add 3%2F2 to both sides of each solution.
highlight_green%28t+=+%283%2Bsqrt%2837%29%29%2F2%29 or highlight_green%28t+=+%283-sqrt%2837%29%29%2F2%29
3) Use the formula for the height of an object propelled upward with an initial velocity of v%5B0%5D ft/sec from an initial height of h%5B0%5D feet. G+=+32+ft%2F%28sec%29%5E2:
h%28t%29+=+-%281%2F2%29Gt%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D Substitute v%5B0%5D+=+64 and h%5B0%5D+=+80
highlight%28h%28t%29+=+-16t%5E2%2B64t%2B80%29 First find the time t at which the object reaches its maximum height. Remember that this equation, when graphed, will show a parabola that opens downward so that the vertex of the graph will be the maximum point of the curve or the time of maximum height, t.
The vertex of this parabola is given by t+=+-b%2F2a where a = -16 and b = 64, so...
t+=+-%2864%29%2F2%28-16%29
t+=+64%2F32
highlight%28t+=2%29seconds.
The object will attain its maximum height in 2 seconds. To find the maximum height of the object, subtitute t = 2 into the beginning equation highlight%28h%28t%29%29and solve for h.
h%282%29+=+-16%282%29%5E2%2B64%282%29%2B80
h%282%29+=+-64%2B128%2B80
highlight%28h%282%29+=+144%29feet is the maximum height attained by the object..