Asked by Layla
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]
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]
Answers
Answered by
Damon
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)
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)
Answered by
Anonymous
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
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
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