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Re: The equation: 2x^2 - 4x = t has no real roots and t is a constant.. [#permalink]
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]
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]
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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