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I know the answer is no but whats the reasoning behind it?

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    For real values of k, Descartes' rule of signs can be used. https://en.wikipedia.org/wiki/Descartes'_rule_of_signs – 2017-02-18

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Hint. One may observe that $$ \begin{align} 2x^2 - kx + k - 2 &=2\left[x^2 - \frac{k}2 x + \frac{k}2 - 1 \right] \\&=2\left[\left(x-\frac{k}4 \right)^2 - \frac{(k-4)^2}{16} \right] \\&=2\left(x-\frac{k-2}2 \right)(x-1) \end{align} $$ Can you finish it?

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For which we need $k<0$ and $k-2>0$, which is impossible.

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    Excellent hint...but perhaps it needs the word "hint" to be added at the beginning, as it is not that straightforward. +1 – 2017-02-18
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$$2x^2 - kx + (k - 2) = 0$$

$$x=\frac{k ±\sqrt{k^2-8(k-2)}}{4}$$

$$x=\frac{k ±\sqrt{(k-4)^2}}{4}$$

$$x=\frac{k ±(k-4)}{4}$$

$$x=1,\frac{(k-2)}{2}$$

So, one root is positive and other root depends on value of $k$, ie, if $k<2$, then one root can be negative.

Both roots can't be negative as one root is positive irrespective of value of k.

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    But for the fact that it gives one *whole* way to reach the solution, this was already answered *this way* by Olivier ... – 2017-02-18
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    @DonAntonio I did it by using the quadratic formula, Olivier used completing the square method. Slightly different. I just had to plug in the values and voila. – 2017-02-18
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You want negative roots. So you want real roots. That means the discriminant

$k^2 - 8(k-2) = k^2 - 8k +16 = (k-4)^2 \ge 0 \Rightarrow k \ge 4.$

We know that sum of roots is equal to $\frac k2 \ge 2$. Hence it is impossible for both roots to be negative.