The sum of all integral values

Question:

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

Solution:

$\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$

$x \in R-\{1,2\}$

$\Rightarrow \mathrm{k}(2 \mathrm{x}-4-\mathrm{x}+1)=2\left(\mathrm{x}^{2}-3 \mathrm{x}+2\right)$

$\Rightarrow \mathrm{k}(\mathrm{x}-3)=2\left(\mathrm{x}^{2}-3 \mathrm{x}+2\right)$

for $x \neq 3, \quad k=2\left(x-3+\frac{2}{x-3}+3\right)$

$x-3+\frac{2}{x-3} \geq 2 \sqrt{2}, \forall x>3$

$\& x-3+\frac{2}{x-3} \leq-2 \sqrt{2}, \forall x<-3$

$\Rightarrow 2\left(x-3+\frac{2}{x-3}+3\right) \in(-\infty, 6-4 \sqrt{2}] \cup[6+4 \sqrt{2}, \infty)$

for no real roots

$\mathrm{k} \in(6-4 \sqrt{2}, 6+4 \sqrt{2})-\{0\}$

Integral $\mathrm{k} \in\{1,2 \ldots . .11\}$

Sum of $\mathrm{k}=66$

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