Solve $ x^2 + \bar{2}x - \bar{2} = \bar{0} $ in $\mathbb{Z}_{11}$

Question

Solve $ x^2 + \bar{2}x - \bar{2} = \bar{0} $ in $\mathbb{Z}_{11}$

I could try plugging in $\bar{0}, \bar{1}, ..., \bar{10}$ and see if I get a solution, however, this method seems too slow and "brute force". Are there other methods for solving this equation?

Can I use the quadratic formula to solve this? Anyway, using the quadratic formula you get the term

$$\sqrt{-\bar{4}}$$

And I'm not sure how to interpret this.

Answer 1

$\!\bmod 11\!:\ $ the discriminant is $\,\color{#c00}{\sqrt{12}}\equiv \sqrt{1}\equiv 1\equiv \color{#0a0}{12},\ $ hence the

quadratic formula $\, \Rightarrow\, x \equiv \dfrac{-2\pm\color{#c00}{\sqrt{12}}}2\,\equiv\, \dfrac{-2\pm \color{#0a0}{12}}2\equiv -1\pm 6\equiv 5,4,\,$ by $\, -7\equiv 4$

Answer 2

Modulo $11$, we have: $$ x^2 + 2x - 2 = x^2 + 2x + 1 - 3 = (x + 1)^2 - 3. $$ Now, does $3$ have any square roots in $\mathbb{Z}_{11}$? Yes: $$ 5^2 = 25 \equiv 3 \mbox{ mod } 11. $$ So, $$ (x + 1)^2 - 5^2, $$ a difference of two squares.