Question

Jacob throws an acorn into the air. It lands in front of him. The acorn's path isdescribed by the equation y=-2x2 + 8x + 6, where x is the acorn's horizontaldistance from him and y is the height of the acorn.Solve -2x2 + 8x + 6 = 0 to see where the acorn hits the ground.Are both solutions reasonable in this situation?O A. x = 2 + 17 = 4.65 feet and x = 2 - 173-0.65 feet. Both solutionsare reasonable.B. x= -8 +52-5.35 feet and x = -8-/7--10.65 feet. Distancecan't be negative, so neither solution is reasonable.C. x = -8 + 7 = -5.35 feet and x = -8 -172-10.65 feet. Bothsolutions are reasonable.O D. x = 2 + 17 - 4.65 feet and x = 2-5-0.65 feet. Distanchpan't benegative, so x = 2 - 7 is not a reasonable solution.

Answer 1

The given equation is:

`y=-2x^2+8x+6`

Where x is the acorn's horizontal distance from him and y is the height of the acorn.

We need to solve the equation to find the roots, as follows:

`\begin{gathered} -2x^2+8x+6=0 \\ \text{ It is written in the form} \\ ax^2+bx+c=0 \\ Where\text{ a=-2, b=8, c=6} \end{gathered}`

Now,we can apply the quadratic formula to solve for x:

`\begin{gathered} x=(-b\pm√(b^2-4ac))/(2a) \\ x=(-8\pm√(8^2-4(-2)(6)))/(2(-2)) \\ x=(-8\pm√(64+48))/(-4) \\ x=(-8\pm√(112))/(-4) \\ x=(-8\pm√(16*7))/(-4) \\ x=(-8\pm4√(7))/(-4) \\ Then\text{ x can be:} \\ x_1=(-8+4√(7))/(-4)=(-8)/(-4)+(4√(7))/(-4)=2-√(7)\text{ And} \\ x_2=(-8-4√(7))/(-4)=(-8)/(-4)-(4√(7))/(-4)=2+√(7) \\ \text{ So:} \\ x=2-√(7)\approx-0.65\text{ AND} \\ x=2+√(7)\approx4.65 \end{gathered}`

The two solutions for x are -0.65 and 4.65.

The distance can't be negative so x=2-square root(7)=-0.65 feet is not a reasonable solution.

The answer is option D