(a) Factoring the left side of $x^{2}+2 x-8=0$, we obtain $(x+4)(x-2)=0$.

Therefore, the roots of $x^{2}+2 x-8=0$ are $x=-4$ and $x=2$.

(b) Since the parabola with equation $y=x^{2}+b x+c$ passes through the points $(1,2)$ and $(2,0)$, then the coordinates of these points satisfy the equation of the parabola.

Therefore, $2=1^{2}+b \cdot 1+c$ (which simplifies to $b+c=1$ ) and $0=2^{2}+2 b+c$ (which simplifies to $2 b+c=-4$ ).

Subtracting the first of these equations from the second, we obtain $(2 b+c)-(b+c)=-4-1$ or $b=-5$.

Since $b=-5$ and $b+c=1$, then $c=1-b=1-(-5)=6$.

Therefore, $b=-5$ and $c=6$.

(We can check that the points $(1,2)$ and $(2,0)$ lie on the parabola with equation $y=x^{2}-5 x+6$.)

(c) Since the point $(0,2)$ lies on the parabola with equation $y=a(x-1)^{2}+\frac{8}{3}$, then $2=a(-1)^{2}+\frac{8}{3}$ or $2=a+\frac{8}{3}$ and so $a=2-\frac{8}{3}=-\frac{2}{3}$.

Thus, the parabola has equation $y=-\frac{2}{3}(x-1)^{2}+\frac{8}{3}$.

We want to determine the value of $d>0$ for which the point $(d, 0)$ lies on this parabola. Substituting, we obtain the following equivalent equations:

$$
\begin{aligned}
0 & =-\frac{2}{3}(d-1)^{2}+\frac{8}{3} \\
\frac{2}{3}(d-1)^{2} & =\frac{8}{3} \\
(d-1)^{2} & =4 \\
d-1 & = \pm 2
\end{aligned}
$$

Therefore, $d=-2+1=-1$ or $d=2+1=3$.

Since $d>0$, then $d=3$.
(a) Determine all real numbers $x$ for which $x^{2}+2 x-8=0$.

(b) Determine the values of $b$ and $c$ for which the parabola with equation $y=x^{2}+b x+c$ passes through the points $(1,2)$ and $(2,0)$.

(c) A ball is thrown from a window at the point $(0,2)$. The ball flies through the air along the parabola with equation $y=a(x-1)^{2}+\frac{8}{3}$, for some real number $a$. Determine the positive real number $d$ for which the ball hits the ground at $(d, 0)$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ff0a25f37b4b6ded9838g-3.jpg?height=517&width=572&top_left_y=1861&top_left_x=1161)