+539 Find all values of t such that floor(t) = 3t + 4 - t^2. If you find more than one value, then list the values you find in increasing order, separated by commas.
+8 We are tasked with finding all values of \( t \) such that:
$$ \lfloor t \rfloor = 3t + 4 - t^2. $$
The floor function \( \lfloor t \rfloor \) represents the greatest integer less than or equal to \( t \). Let \( n = \lfloor t \rfloor \), so that \( n \) is an integer and:
$$ n \leq t < n + 1. $$
Therefore, we can rewrite the given equation as:
$$ n = 3t + 4 - t^2. $$
Rearranging the equation, we get the quadratic equation in \( t \):
$$ t^2 - 3t + (n - 4) = 0. $$
Using the quadratic formula to solve for \( t \), we get:
$$ t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(n - 4)}}{2(1)} = \frac{3 \pm \sqrt{9 - 4(n - 4)}}{2} = \frac{3 \pm \sqrt{25 - 4n}}{2}. $$
Therefore, the solutions for \( t \) are:
$$ t = \frac{3 \pm \sqrt{25 - 4n}}{2}. $$
For real solutions to exist, the expression under the square root must be non-negative, i.e.,
$$ 25 - 4n \geq 0, $$
which simplifies to:
$$ n \leq \frac{25}{4} = 6.25. $$
Since \( n \) is an integer, the largest possible value of \( n \) is 6. Thus, \( n \) can range from \( -\infty \) to 6.
Now we solve for \( t \) for each integer value of \( n \) from \( n = 6 \) downward, checking whether each value of \( t \) lies within the interval \( n \leq t < n+1 \).
For \( n = 6 \):
The equation becomes:
$$ t^2 - 3t + 2 = 0. $$
Solving for \( t \) using the quadratic formula:
$$ t = \frac{3 \pm \sqrt{9 - 8}}{2} = \frac{3 \pm \sqrt{1}}{2} = \frac{3 \pm 1}{2}. $$
Thus, the solutions are:
$$ t = \frac{3 + 1}{2} = 2 \quad \text{or} \quad t = \frac{3 - 1}{2} = 1. $$
Check if these values satisfy \( 6 \leq t < 7 \):
Only \( t = 2 \) satisfies this condition, so \( t = 2 \) is a valid solution for \( n = 6 \).
For \( n = 5 \):
The equation becomes:
$$ t^2 - 3t + 1 = 0. $$
Solving for \( t \):
$$ t = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}. $$
The solutions are irrational, so we check whether these values satisfy \( 5 \leq t < 6 \). They do not, so there are no valid solutions for \( n = 5 \).
For \( n = 4 \):
The equation becomes:
$$ t^2 - 3t = 0. $$
Factoring:
$$ t(t - 3) = 0. $$
Thus, \( t = 0 \) or \( t = 3 \). Check whether these satisfy \( 4 \leq t < 5 \):
Neither \( t = 0 \) nor \( t = 3 \) lies within the interval \( 4 \leq t < 5 \), so there are no valid solutions for \( n = 4 \).
For \( n = 3 \):
The equation becomes:
$$ t^2 - 3t - 1 = 0. $$
Solving for \( t \):
$$ t = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}. $$
The solutions are irrational, so we check whether these values satisfy \( 3 \leq t < 4 \). They do not, so there are no valid solutions for \( n = 3 \).
For \( n = 2 \):
The equation becomes:
$$ t^2 - 3t - 2 = 0. $$
Solving for \( t \):
$$ t = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}. $$
The solutions are irrational, so we check whether these values satisfy \( 2 \leq t < 3 \). They do not, so there are no valid solutions for \( n = 2 \).
For \( n = 1 \):
The equation becomes:
$$ t^2 - 3t - 3 = 0. $$
Solving for \( t \):
$$ t = \frac{3 \pm \sqrt{9 + 12}}{2} = \frac{3 \pm \sqrt{21}}{2}. $$
The solutions are irrational, so we check whether these values satisfy \( 1 \leq t < 2 \). They do not, so there are no valid solutions for \( n = 1 \).
For \( n = 0 \):
The equation becomes:
$$ t^2 - 3t - 4 = 0. $$
Solving for \( t \):
$$ t = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}. $$
Thus, the solutions are:
$$ t = \frac{3 + 5}{2} = 4 \quad \text{or} \quad t = \frac{3 - 5}{2} = -1. $$
Check if these satisfy \( 0 \leq t < 1 \):
Only \( t = -1 \) satisfies this condition, so \( t = -1 \) is a valid solution for \( n = 0 \).
For \( n = -1 \):
The equation becomes:
$$ t^2 - 3t - 5 = 0. $$
Solving for \( t \):
$$ t = \frac{3 \pm \sqrt{9 + 20}}{2} = \frac{3 \pm \sqrt{29}}{2}. $$
The solutions are irrational, so we check whether these values satisfy \( -1 \leq t < 0 \). They do not, so there are no valid solutions for \( n = -1 \).
Conclusion:
The only valid solution is \( t = -1 \).
