McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
3. The Quadratic Formula and the Discriminant
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Exercise 9 Page 194

Use the Quadratic Formula.

0.78 seconds

Practice makes perfect

We are given the equation that models the height of the ride. h(t)=- 16t^2-64t+60 To find the time it takes to drop from the initial 60 feet to ground level, we need to find the time when the height is 0. We can do this by solving a quadratic equation. - 16t^2-64t+60=0Let's use the Quadratic Formula. 2 &Quadratic equation: && ax^2+ bx+ c=0 &Solutions:&&x=- b±sqrt(b^2-4 a c)/2 a Let's identify the coefficients in our quadratic expression. - 16t^2-64t+60= - 16t^2+( - 64)t+ 60 We see that a= - 16, b= - 64, and c= 60. Let's substitute these values into the Quadratic Formula.

t=- b±sqrt(b^2-4ac)/2a
t=- ( -64)±sqrt(( - 64)^2-4( -16)( 60))/2( -16)
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Simplify right-hand side
t=64±sqrt(( - 64)^2-4( -16)( 60))/2( -16)
t=64±sqrt(4096-4( -16)( 60))/2( -16)
t=64±sqrt(4096-240( -16))/2( -16)
t=64±sqrt(4096-(-3840))/-32
t=64±sqrt(4096+3840)/-32
t=64±sqrt(7936)/-32

Let's use a calculator to find approximate values of these solutions. 64+sqrt(7936)/-32&≈ -4.78 64-sqrt(7936)/-32&≈ 0.78 Since the solution we are looking for represents time, we need the positive solution.



It takes approximately 0.78 seconds for the riders to drop from the initial 60 feet to ground level.