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The equation: 2x^2 - 4x = t has no real roots and t is a constant..
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21 Sep 2021, 07:35
21 Sep 2021, 07:35
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Question Stats:
50% (01:02) correct
50% (01:11) wrong based on 22 sessions
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The equation: \(2x^2 - 4x = t\) has no real roots and \(t\) is a constant.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
\(t\) |
-3 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
The equation: 2x^2 - 4x = t has no real roots and t is a constant..
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Updated on: 23 Sep 2021, 12:29
Updated on: 23 Sep 2021, 12:29
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KarunMendiratta wrote:
The equation: \(2x^2 - 4x = t\) has no real roots and \(t\) is a constant.
Re-arrange the equation to \(2x^2 - 4x - t = 0\)
and if there are no real roots then for an equation \(ax^2 + bx+c = 0\) then \( b^2 - 4ac < 0\)
so \((-4)^2 - 4*2*(-t) < 0 => t < -2\)
Which means, it can any value below 2 , so Answer is D).
Re: The equation: 2x^2 - 4x = t has no real roots and t is a constant..
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23 Sep 2021, 12:07
23 Sep 2021, 12:07
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eskay1981 wrote:
KarunMendiratta wrote:
The equation: \(2x^2 - 4x = t\) has no real roots and \(t\) is a constant.
Re-arrange the equation to \(2x^2 - 4x - t = 0\)
and if there are no real roots then for an equation \(ax^2 + bx+c = 0\) then \( b^2 - 4ac < 0\)
so \((-4)^2 - 4*2*(-t) < 0 => t < 2\)
Which means, it can any value below 2 , so Answer is D).
I think the condition is \( t < -2\) (the answer is still D). Note that if \( t < 2 \) we can set \(t = 0\) and \(2x^2 - 4x =0\) clearly has 0 as a solution.
