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16 Jan 2021, 20:17
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
56% (02:21) correct
44%
(02:11)
wrong
based on 212
sessions
History
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For which values of \(t\), the parabola, \(y = tx^2 + 16tx + 68t\), will not lie above X axis?
A. \(t < 0\)
B. \(0 < t < 4\)
C. \(t < -4\)
D. \(t < -4\) or \(t > 4\)
E. \(-4 < t < 0\)
A. \(t < 0\)
B. \(0 < t < 4\)
C. \(t < -4\)
D. \(t < -4\) or \(t > 4\)
E. \(-4 < t < 0\)
16 Jan 2021, 23:46
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y = tx^2 + 16xt + 68t
y = t (x^2 + 16x + 68)
Discriminant (D) of the quadratic equation x^2 + 16x + 68 = 16^2 - 4*1*68 = -16 (negative) that means roots are imaginary. In other words the parabola is not touching the x-axis.
Hence when t is positive, the parabola is upward whereas when t is negative, the parabola is downward
IMO A
y = t (x^2 + 16x + 68)
Discriminant (D) of the quadratic equation x^2 + 16x + 68 = 16^2 - 4*1*68 = -16 (negative) that means roots are imaginary. In other words the parabola is not touching the x-axis.
Hence when t is positive, the parabola is upward whereas when t is negative, the parabola is downward
IMO A
General Discussion
11 Apr 2025, 03:58
Kudos
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First, let me factor out t from the equation:
y = t(x2 + 16x + 68)
For this parabola to not lie above the X-axis, we need y ≤ 0 for some value of x, which means:
t(x2 + 16x + 68) ≤ 0
This inequality is satisfied in two cases:
t > 0 and (x2 + 16x + 68) ≤ 0, or
t < 0 and (x2 + 16x + 68) ≥ 0
Let's analyze the quadratic expression x2 + 16x + 68:
This has discriminant b2 - 4ac = 162 - 4(1)(68) = 256 - 272 = -16
Since the discriminant is negative, the quadratic expression has no real roots
Since the leading coefficient is positive, the expression x2 + 16x + 68 > 0 for all x
Therefore:
If t > 0, the inequality t(x2 + 16x + 68) ≤ 0 is never satisfied (since x2 + 16x + 68 > 0)
If t < 0, the inequality t(x2 + 16x + 68) ≤ 0 is always satisfied (since x2 + 16x + 68 > 0)
If t = 0, then y = 0 for all x, so the parabola lies on the X-axis
So the parabola will not lie above the X-axis when t < 0.
The answer is A. t < 0
y = t(x2 + 16x + 68)
For this parabola to not lie above the X-axis, we need y ≤ 0 for some value of x, which means:
t(x2 + 16x + 68) ≤ 0
This inequality is satisfied in two cases:
t > 0 and (x2 + 16x + 68) ≤ 0, or
t < 0 and (x2 + 16x + 68) ≥ 0
Let's analyze the quadratic expression x2 + 16x + 68:
This has discriminant b2 - 4ac = 162 - 4(1)(68) = 256 - 272 = -16
Since the discriminant is negative, the quadratic expression has no real roots
Since the leading coefficient is positive, the expression x2 + 16x + 68 > 0 for all x
Therefore:
If t > 0, the inequality t(x2 + 16x + 68) ≤ 0 is never satisfied (since x2 + 16x + 68 > 0)
If t < 0, the inequality t(x2 + 16x + 68) ≤ 0 is always satisfied (since x2 + 16x + 68 > 0)
If t = 0, then y = 0 for all x, so the parabola lies on the X-axis
So the parabola will not lie above the X-axis when t < 0.
The answer is A. t < 0
