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The equation: 2x^2 - 4x = t has no real roots and t is a constant..

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KarunMendiratta
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The equation: 2x^2 - 4x = t has no real roots and t is a constant.. [#permalink] New post   21 Sep 2021, 07:35
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The equation: \(2x^2 - 4x = t\) has no real roots and \(t\) is a constant.

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
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D
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eskay1981
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The equation: 2x^2 - 4x = t has no real roots and t is a constant.. [#permalink] New post   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).

Originally posted by eskay1981 on 21 Sep 2021, 08:08.
Last edited by eskay1981 on 23 Sep 2021, 12:29, edited 1 time in total.
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Re: The equation: 2x^2 - 4x = t has no real roots and t is a constant.. [#permalink] New post   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.
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eskay1981
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Re: The equation: 2x^2 - 4x = t has no real roots and t is a constant.. [#permalink] New post   23 Sep 2021, 12:28
juangg wrote:
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.


Thanks for pointing it out. I made a typo.
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Re: The equation: 2x^2 - 4x = t has no real roots and t is a constant.. [#permalink] New post   23 Sep 2021, 20:17
juangg wrote:
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.


juangg
Dear, \(t\) cannot be zero as it has to be less than -2
The solution cannot be zero as well, as zero is a real number and the question clearly mentions that the roots are non real.
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Re: The equation: 2x^2 - 4x = t has no real roots and t is a constant.. [#permalink] New post   24 Sep 2021, 05:50
KarunMendiratta wrote:

Dear, \(t\) cannot be zero as it has to be less than -2
The solution cannot be zero as well, as zero is a real number and the question clearly mentions that the roots are non real.


I agree. I was just arguing why \( t < 2 \) could not be the correct restriction.
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Re: The equation: 2x^2 - 4x = t has no real roots and t is a constant.. [#permalink]
24 Sep 2021, 05:50
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