To solve the equation [tex]\((t - 3)^2 = 6\)[/tex] for [tex]\(t\)[/tex], we proceed step-by-step:
1. Take the square root of both sides:
[tex]\[
\sqrt{(t - 3)^2} = \sqrt{6}
\][/tex]
This gives us two possible equations because taking the square root of a number results in both a positive and a negative root:
[tex]\[
t - 3 = \sqrt{6} \quad \text{and} \quad t - 3 = -\sqrt{6}
\][/tex]
2. Solve each equation for [tex]\(t\)[/tex]:
- For the first equation, [tex]\(t - 3 = \sqrt{6}\)[/tex]:
[tex]\[
t = 3 + \sqrt{6}
\][/tex]
By substituting the value of [tex]\(\sqrt{6}\)[/tex], we get:
[tex]\[
t \approx 3 + 2.449489742783178
\][/tex]
Therefore,
[tex]\[
t \approx 5.449489742783178
\][/tex]
- For the second equation, [tex]\(t - 3 = -\sqrt{6}\)[/tex]:
[tex]\[
t = 3 - \sqrt{6}
\][/tex]
By substituting the value of [tex]\(\sqrt{6}\)[/tex], we get:
[tex]\[
t \approx 3 - 2.449489742783178
\][/tex]
Therefore,
[tex]\[
t \approx 0.5505102572168221
\][/tex]
Thus, the values of [tex]\(t\)[/tex] that satisfy the equation [tex]\((t - 3)^2 = 6\)[/tex] are approximately [tex]\(t \approx 5.449489742783178\)[/tex] and [tex]\(t \approx 0.5505102572168221\)[/tex].
So, the arrow is at a height of 48 ft after approximately 0.5505102572168221 seconds and 5.449489742783178 seconds.
