Range of $k$ for which equation has positive roots 
Range of $k$ for which both the roots of the equation $(k-2)x^2+(2k-8)x+3k-17=0$ are positive.

Try: if $\alpha,\beta>0$ be the roots of the equation. Then $$\alpha+\beta=\frac{8-2k}{k-2}>0\Rightarrow k\in(2,4)$$
And $$\frac{3k-17}{k-2}>0\Rightarrow k\in(-\infty,2)\cup \bigg(\frac{17}{3},\infty\bigg)$$
I have got $k=\phi$. I did not understand where i am wrong , please explain, Thanks

Use quadratic formula. $$(k-2)x^2+(2k-8)x+3k-17=0$$ gives $$x=\frac{-(2k-8)\pm\sqrt{(2k-8)^2-4(k-2)(3k-17)}}{2(k-2)}=-\frac{k-4}{k-2}\pm\frac{\sqrt{-8k^2+60k-72}}{2(k-2)}$$ Hence $$x=-1+\frac2{k-2}\pm\frac{\sqrt{(k-6)(3-2k)}}{k-2}$$ Now we need only find when the negative root is positive since the positive one would then be positive as well. 
So $$\frac2{k-2}>1+\frac{\sqrt{(k-6)(3-2k)}}{k-2}\implies \cdots$$
