$$|x+2|+|x-3|=5$$ and
$(x^2-x-6)\leq0$
Have the same real solution of $[-2,3]$
I was wondering if there was a formula to convert from one to the other, such as taking a quadratic inequality and turning it into an absolute values equation. I know it's not very practical as the method for solving both formats is similar, but I was wondering about the relationship between the formats.
$$|x-a|+|x-b|=|a-b|\Leftrightarrow(x-a)(x-b)\leq0$$
You can use even the following equation $$\sqrt{x-x^2}=\sqrt{x-x^2},$$ which gives $[0,1]$.
Also, the equation $$\sqrt{1-x^2-y^2}=\sqrt{1-x^2-y^2}$$ is an equation of the disc.
Also, $$|x+y|+|x-y|=2$$ is an equation of the square.