No real roots?

**acc100jt** · September 14th 2009, 06:02 PM

Determine the possible integer values of n for which there exists some real number k such that the equation $2x^{n}-4x+1=k$ has no real roots.

Anyone can help? Thanks!!

**pickslides** · September 14th 2009, 06:14 PM

For n = 2 use the discriminant <0.

$2x^{2}-4x+1-k=0$

Solve $(-4)^2-4\times 2\times (1-k) < 0$

Maybe this idea is a little simplistic for these problem

**acc100jt** · September 14th 2009, 06:47 PM

obviously n has to be even, because i need the signs of the limits of the quadratic $f(x)=2x^{n}-4x+1$ to be the same, when x tend to both +ve and -ve infinity. Then the graph of f(x) will curve either upwards or downwards at both ends, so there will exists some k where the equation has no solution. But how to give a formal solution??

or is there any other way to solve it???

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