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Roots and inequalities

The roots or the zeros of a numerical function $$f$$ form the set of values that solve the equation $$f(x)=0$$.

For instance, the roots of the second order expression $$$f(x) = ax^2 + bx+c$$$ with $$a\ne 0$$ depend on the sign of the discriminant $$$\Delta = b^2 - 4 ac.$$$

  • If $$\Delta\lt 0$$, $$f$$ has no real root.
  • If $$\Delta = 0$$, $$f$$ has a unique zero $$-b / (2a)$$, called a repeated root .
  • If $$\Delta\gt 0$$, $$f$$ has two distinct real roots $$$\displaystyle x_- = \frac{-b-\sqrt{\Delta}}{2a},\quad x_+ = \frac{-b+\sqrt{\Delta}}{2a}.$$$

  • $$x^2+1=0$$ has no real root, because $$\Delta = -4\lt 0 $$.
  • $$x^2-2x+1=0$$ has a repeated root $$x=1$$, because $$\Delta = 0 $$.
  • $$x^2+2x=0$$ has a two real roots, because $$\Delta = 4 \gt 0$$. The roots are $$x_- = (-2-2)/2 = -2$$ and $$x_+=(-2+2)/2 = 0$$.

Inequalities involving numerical functions can be rewritten as $$$f(x)\lt 0,\qquad f(x)\le 0$$$

The inequality $$g(x)\gt 1$$ can be written as $$1-g(x)\lt 0$$.

For a quadratic function with $$a\gt 0$$ and discriminant $$\Delta = b^2-4ac\gt0$$, the roots are $$\displaystyle x_\pm = \frac{-b\pm\sqrt{\Delta}}{2a}$$. So, we have $$$ f(x) = ax^2 + bx + c = a(x-x_-)(x-x_+).$$$ So $$f\lt 0$$ on $$(x_-,x_+)$$ and $$f \ge 0$$ on $$(-\infty,x_-]\cup[x_+,+\infty)$$.

To compute the sign of a function, we decompose it as the product or ratio of simple functions. Take $$f=g/h$$, with $$g$$ and $$h$$ having distinct non stationary zeros (if $$g(x)=0$$ then $$g'(x)\ne0$$). Then, $$f$$ alternates sign at the roots of $$g$$ and $$h$$. Its limit at $$\pm\infty$$ gives the sign at $$\pm\infty$$.

$$f(x)=(x^2-1)/x$$ changes sign at $$-1$$, $$0$$ and $$1$$. Its limit at $$-\infty$$ is $$-1$$. Hence it is strictly negative on $$(-\infty,-1)\cup(0,1)$$.

Sign of the function $$f(x) = \frac{x^2-1}{x}$$
Sign of the function $$f(x) = \frac{x^2-1}{x}$$