## College Algebra 7th Edition

$\displaystyle \frac{1}{x-1}-\frac{2}{x^{2}}=0$ $\displaystyle \frac{1}{x-1}=\frac{2}{x^{2}}$ We cross-multiply: $x^{2}=2(x-1)$ $x^{2}-2(x-1)=0$ $x^{2}-2x+2=0$ Next we use the quadratic formula: $\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^{2}-4*1*2}}{2(1)}=\frac{2\pm\sqrt{4-8}}{2}=\frac{2\pm\sqrt{-4}}{2}$ Since the discriminant is negative, there are no real solutions.