Question

Jacob throws an acorn into the air. It lands in front of him. The acorn's path is

Answer 1

The acorn hits the ground at x = 1.232 and x = -2.232

Also, only one solution is reasonable which is x = 1.232 since the acorn lands in front of him.

To answer the question, we need to solve the quadratic equation.

How to solve the quadratic equation?

Since the acorn's path is described by y = -3x² + 6x + 6, where x is the acorn's horizontal distance from him and y is the height of the acorn.

We solve -3x² + 6x + 6 = 0 to see where the acorn hits the ground.

So, -3x² + 6x + 6 = 0

Using the quadratic formula, we have

$x = \frac{-(-3) +/- \sqrt{6^{2} - 4X(-3)X6} }{2 X (-3)} \\x = \frac{3 +/- \sqrt{36 + 72} }{-6} \\x = \frac{3 +/- \sqrt{108} }{-6} \\x = \frac{3 +/- 10.39 }{-6} \\x = \frac{3 - 10.39 }{-6} or x = \frac{3 + 10.39 }{-6} \\x = \frac{-7.39 }{-6} or x = \frac{13.39 }{-6} \\x = 1.232 or x = -2.232$

So, the acorn hits the ground at x = 1.232 and x = -2.232

Also, only one solution is reasonable which is x = 1.232 since the acorn lands in front of him.

Learn more about quadratic equation here:

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Answer 2

Answer:

• hits the ground at x = -0.732, and x = 2.732
• only the positive solution is reasonable

Step-by-step explanation:

The acorn will hit the ground where the value of x is such that y=0. We can find these values of x by solving the quadratic using any of several means.

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graphing

The attachment shows a graphing calculator solution to the equation

  -3x^2 + 6x + 6 = 0

The values of x are -0.732 and 2.732. The negative value is the point where the acorn would have originated from if its parabolic path were extrapolated backward in time. Only the positive horizontal distance is a reasonable solution.

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completing the square

We can also solve the equation algebraically. One of the simplest methods is "completing the square."

  -3x^2 +6x +6 = 0

  x^2 -2x = 2 . . . . . . . . divide by -3 and add 2

  x^2 -2x +1 = 2 +1 . . . . add 1 to complete the square

  (x -1)^2 = 3 . . . . . . . . written as a square

  x -1 = ±√3  . . . . . . . take the square root

  x = 1 ±√3 . . . . . . . add 1; where the acorn hits the ground

The numerical values of these solutions are approximately ...

  x ≈ {-0.732, 2.732}

The solutions to the equation say the acorn hits the ground at a distance of -0.732 behind Jacob, and at a distance of 2.732 in front of Jacob. The "behind" distance represents and extrapolation of the acorn's path backward in time before Jacob threw it. Only the positive solution is reasonable.