Given equation

We need integral values of $k$ with $k \ne 0$ such that $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$ has no real roots in $x$.

Simplify the left-hand side

Cross-multiply

Since $x\ne 1,2$ and $k\ne 0$, we can cross-multiply: $k(x-3)=2(x-1)(x-2)$

Expand: $kx-3k=2(x^2-3x+2)$ $kx-3k=2x^2-6x+4$

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

Condition for no real roots

For the quadratic in $x$ $2x^2-(k+6)x+(3k+4)=0$ to have no real roots, its discriminant must be negative: $D=(k+6)^2-4\cdot 2\cdot (3k+4)<0$

Compute: $D=k^2+12k+36-24k-32$ $D=k^2-12k+4<0$

Solve the inequality

$k^2-12k+4<0$

Roots of $k^2-12k+4=0$ are $k=\frac{12\pm\sqrt{144-16}}{2}=\frac{12\pm\sqrt{128}}{2}=6\pm 4\sqrt{2}$

Hence, $6-4\sqrt{2}<k<6+4\sqrt{2}$

Numerically, $6-4\sqrt{2}\approx 0.343, \qquad 6+4\sqrt{2}\approx 11.657$

Therefore the integral values of $k$ are $k=1,2,3,4,5,6,7,8,9,10,11$

(and $k\ne 0$ is already satisfied).

Check excluded values $x=1,2$

The original equation is undefined at $x=1,2$.

Substitute into the quadratic:

• At $x=1$: $2-(k+6)+(3k+4)=2k \ne 0$ for $k\ne 0$
• At $x=2$: $8-2(k+6)+(3k+4)=k \ne 0$ for $k\ne 0$

So no extraneous issue arises; the quadratic having no real roots is the correct condition.

Sum of all such integral values

$1+2+3+\cdots+11=\frac{11\cdot 12}{2}=66$

Comparison with stored answer

Derived answer = $66$.

This matches the stored correct answer.