The sum of all integral values of $$k$$ ($$k \neq 0$$) for which the equation $$\frac{2}{x-1} - \frac{1}{x-2} = \frac{2}{k}$$ in $$x$$ has no real roots, is _________

We begin with the given equation

$$\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k},\qquad k\neq 0.$$

To clear the fractions we first take the common denominator on the left side. The common denominator of the two fractions is $$(x-1)(x-2).$$ Hence

$$\frac{2}{x-1}-\frac{1}{x-2}\;=\;\frac{2(x-2)-(x-1)}{(x-1)(x-2)}.$$

Expanding the numerator we have

$$2(x-2)-(x-1)=2x-4-x+1=x-3.$$

Therefore the equation becomes

$$\frac{x-3}{(x-1)(x-2)}=\frac{2}{k}.$$

Now we cross-multiply (which is allowed because none of the denominators is zero when a solution is valid):

$$k(x-3)=2\,(x-1)(x-2).$$

Next we expand the right-hand side. First note that

$$(x-1)(x-2)=x^2-3x+2.$$

Multiplying by the factor $$2$$ we get

$$2(x-1)(x-2)=2x^2-6x+4.$$

So the cross-multiplied equation is

$$k(x-3)=2x^2-6x+4.$$

We now bring every term to one side to form a quadratic equation in $$x$$. Distributing $$k$$ on the left gives

$$kx-3k=2x^2-6x+4.$$

Subtracting the left-hand side from both sides gives

$$0=2x^2-6x+4-kx+3k.$$

Grouping like terms in $$x$$ we write

$$0=2x^2-(k+6)x+(3k+4).$$

Thus the quadratic in standard form is

$$2x^2-(k+6)x+(3k+4)=0.$$

The quadratic formula tells us that a quadratic equation $$ax^2+bx+c=0$$ has real roots precisely when its discriminant $$\Delta=b^2-4ac$$ is non-negative. We want no real roots, so we require $$\Delta\lt 0.$$

Here $$a=2,\; b=-(k+6),\; c=3k+4.$$ Hence

$$\Delta=(-(k+6))^2-4\cdot2\cdot(3k+4).$$

Calculating each part we get

$$(k+6)^2=k^2+12k+36,$$

and

$$4\cdot2\cdot(3k+4)=8(3k+4)=24k+32.$$

Therefore

$$\Delta=k^2+12k+36-(24k+32)=k^2-12k+4.$$

For no real roots we need

$$k^2-12k+4\lt 0.$$

This is a quadratic inequality in $$k$$. Because the coefficient of $$k^2$$ is positive, the expression is negative between its two real roots. We find the roots by setting the quadratic equal to zero:

$$k^2-12k+4=0.$$

Using the quadratic formula with respect to $$k$$ (now $$a=1,\; b=-12,\; c=4$$) we get

$$k=\frac{12\pm\sqrt{(-12)^2-4\cdot1\cdot4}}{2}=\frac{12\pm\sqrt{144-16}}{2}=\frac{12\pm\sqrt{128}}{2}=\frac{12\pm8\sqrt2}{2}=6\pm4\sqrt2.$$

Thus the inequality $$k^2-12k+4\lt 0$$ holds for

$$6-4\sqrt2\lt k\lt 6+4\sqrt2.$$

Since $$4\sqrt2\approx5.656854,$$ the numerical bounds are approximately

$$0.343146\lt k\lt 11.656854.$$

We are asked for integral (integer) values of $$k$$ with the additional condition $$k\neq0$$. The integers that satisfy the inequality are therefore

$$k=1,2,3,4,5,6,7,8,9,10,11.$$

Adding these values we obtain

$$1+2+3+4+5+6+7+8+9+10+11=\frac{11\,(1+11)}{2}=66.$$

So, the answer is $$66$$.