Solve the equation -1/m + 1 - m = 0.

Step-by-step solution

1. 1

   Record the domain. The term $\tfrac1m$ forces $m\neq 0$, so any candidate root of $0$ would have to be discarded.

2. 2

   Multiply through by $m$. Since $m\neq 0$ this is reversible:

   $-1+m-m^2=0$

3. 3

   Normalise the leading coefficient. Multiply by $-1$ so the $m^2$ term is positive:

   $m^2-m+1=0$

4. 4

   Compute the discriminant and read off the conclusion. With $a=1$, $b=-1$, $c=1$:

   $\Delta=(-1)^2-4(1)(1)=1-4=-3<0$

   A negative discriminant means no real solutions. Geometrically, $y=m^2-m+1=\left(m-\tfrac12\right)^2+\tfrac34$ has minimum value $\tfrac34>0$, so its graph never touches the axis.

5. 5

   Give the complex roots for completeness. $\sqrt{-3}=i\sqrt3$, so

   $m=\frac{1\pm i\sqrt3}{2}$

   These are the two primitive sixth roots of unity; each has modulus $1$. Check: their sum is $1$ and their product is $\tfrac{1+3}{4}=1$, matching $-\tfrac{b}{a}$ and $\tfrac{c}{a}$ $\checkmark$.