Did I solve this quadratic inequality correctly? Is my work correct?
9t^2 + 1 ≤ -6t
9t^2 + 1 -(-6t) ≤-6t-(-6t)
9^2 +6t +1≤0
(3t+1)^2≤0
(3t+1≤0), (3t+1≥0),
(t≤-1/3), (t≥-1/3)
t=-1/3
(-∞, -1/3]
9 t^2 + 6 t + 1 </= 0 agree
now y = 9 t^2 + 6 t + 1
Is a parabola with the vertex down (holds water)
It satisfies the condition (y </=0) between the two intercepts with the x axis.
BUT there is only ONE at x = -1/3
thus
ONLY at that point (t = -1/3)
from (3t+1)^2≤0
it should be intuitively obvious that anything squared cannot be less zero
So the only case possible is
3t=0
t=-1/3 , that is, when the equality holds true
To verify if your quadratic inequality solution is correct, let's go through the steps you took:
1. Start with the quadratic inequality: 9t^2 + 1 ≤ -6t
2. Combine like terms: 9t^2 + 6t + 1 ≤ 0
3. Factor the quadratic equation: (3t + 1)(3t + 1) ≤ 0
4. Rewrite the equation as a product: (3t + 1)^2 ≤ 0
5. To determine the solutions, we consider two cases:
a) (3t + 1) ≤ 0
b) (3t + 1) ≥ 0
Solving for case a):
(3t + 1) ≤ 0
3t ≤ -1
t ≤ -1/3
Solving for case b):
(3t + 1) ≥ 0
3t ≥ -1
t ≥ -1/3
6. Combining the solutions from both cases, the final solution set is: t ≤ -1/3 or t ≥ -1/3. This can be written as: (-∞, -1/3] U [-1/3, ∞)
Based on the steps you followed, your solution and work appear to be correct.