In my draft on the Unified Substitution Method for Integration (USM), I ended up using a very systematic normalization of quadratics and radicals: you shift and scale so that everything happens in the nice, symmetric interval $[-1,1]$ or outside it.
That same normalization idea leads to a neat side trick for solving inequalities. This post presents that trick, shows it on two main examples, and compares it to the traditional methods.
We’ll use three inequalities:
1. A rational inequality:
$$\frac{x-2}{x+3} \ge 0$$
2. A quadratic that changes sign:
$$x^2 - 4x + 3 \le 0$$
3. A quadratic that is always positive:
$$x^2 - 2x + 3 \ge 0$$
This is basically the same normalization that USM uses for integrals, but repurposed for inequalities.
The core idea: normalize to $[-1,1]$
The general pattern is:- Rewrite the expression so that it involves
$$\frac{(x+b)-a}{(x+b)+a}
\quad\text{or}\quad
(x+b)^2 \pm a^2$$
with $a>0$.
- Notice that
$$(x+b)-a = a\Big(\frac{x+b}{a}-1\Big),\quad
(x+b)+a = a\Big(\frac{x+b}{a}+1\Big),$$
so the sign is controlled by how $\dfrac{x+b}{a}$ compares to $-1$ and $1$.
- Reduce the problem to very simple linear inequalities in $\dfrac{x+b}{a}$.
Example 1 – Rational inequality
$$\frac{x-2}{x+3} \ge 0$$
Traditional method (very briefly)
Zeros and pole:$x-2=0 \Rightarrow x=2$,
$x+3=0 \Rightarrow x=-3$.
A sign chart on the intervals $(-\infty,-3)$, $(-3,2)$, $(2,\infty)$ shows the fraction is $\ge 0$ on
$$(-\infty,-3)\cup[2,\infty).$$
Normalization trick
Write$$\frac{x-2}{x+3} = \frac{x+\color{red}{b-a}}{x+\color{red}{b+a}}$$
and solve this little system:
$$\left. b-a=-2\atop a+b=3\right\}$$
This gives:
$$a = \frac{5}{2},\quad b = \frac{1}{2}.$$
Now plug $a=\dfrac52$, $b=\dfrac12$:
- $\dfrac{x+b}{a} \ge 1 \Rightarrow x \ge 2$,
- $\dfrac{x+b}{a} \le -1 \Rightarrow x \le -3$, but $x=-3$ is excluded.
$$(-\infty,-3)\cup[2,\infty).$$
No sign chart; just a tiny $2\times 2$ system and linear inequalities.
Example 2 – Quadratic inequality
$$x^2 - 11x + 13 \ge 0$$
Traditional method (very briefly)
Compute the discriminant:
$$\Delta = 11^2 - 4\cdot 1 \cdot 13 = 121 - 52 = 69 > 0.$$
Roots:
$$x = \frac{11 \pm \sqrt{69}}{2}.$$
Since the parabola opens upwards, the quadratic is $\ge 0$ outside the interval between the roots:
$$x \in \left(-\infty,\frac{11-\sqrt{69}}{2}\right] \cup \left[\frac{11+\sqrt{69}}{2},\infty\right).$$
Normalization trick (completing the square)
Complete the square:
$$x^2 - 11x + 13 = \left(x - \frac{11}{2}\right)^2 - \frac{69}{4}.$$
This fits
$$(x+\color{red}{b})^2 - \color{red}{a^2}$$
with
$$b = -\frac{11}{2}, \quad a = \frac{\sqrt{69}}{2}.$$
Again, we have:
$$\frac{x+b}{a} \le -1 \quad\text{or}\quad \frac{x+b}{a} \ge 1.$$
Substituting $b=-\dfrac{11}{2}$, $a=\dfrac{\sqrt{69}}{2}$:
$$x \le \frac{11-\sqrt{69}}{2}\quad\text{or}\quad x \ge \frac{11+\sqrt{69}}{2}.$$
So again,
$$x \in \left(-\infty,\frac{11-\sqrt{69}}{2}\right] \cup \left[\frac{11+\sqrt{69}}{2},\infty\right).$$
Example 3 – Rational inequality with quadratic numerator and denominator
$$\frac{x^2+3x+2}{x^2-3x-18} \ge 0$$
Normalization trick
First complete the square in numerator and denominator.
For the numerator:
$$x^2 + 3x + 2= \left(x + \frac{3}{2}\right)^2 - \frac{1}{4},$$
so it has the form
$$(x + b_1)^2 - a_1^2$$
with
$$b_1 = \frac{3}{2}, \quad a_1 = \frac{1}{2}.$$
Solve for
$$\left|\frac{x + \frac{3}{2}}{\frac{1}{2}}\right| \ge 1,$$which is
$$\left|2x + 3\right|\ge 1\;\Longleftrightarrow\;x \le -2 \quad\text{or}\quad x \ge -1.$$
Similarly,
$$x^2 + 3x + 2 \le 0\;\Longleftrightarrow\;-2 \le x \le -1.$$
So the numerator is
$\ge 0\, \text{on } (-\infty,-2] \cup [-1,\infty),$
$\le 0\, \text{on } [-2,-1].$
For the denominator:
$$x^2 - 3x - 18= \left(x - \frac{3}{2}\right)^2 - \frac{81}{4},$$
so this is
$$(x + b_2)^2 - a_2^2$$
with
$$b_2 = -\frac{3}{2}, \quad a_2 = \frac{9}{2}.$$
Then we have
$$\left|\frac{x - \frac{3}{2}}{\frac{9}{2}}\right|\ge 1,$$i.e.
$$x \le -3 \quad\text{or}\quad x \ge 6.$$Likewise,
$$x^2 - 3x - 18 \le 0\;\Longleftrightarrow\;-3 \le x \le 6.$$
Note that $x=-3$ and $x=6$ make the denominator zero, so they must be excluded from the domain.
So the denominator is
$> 0\, \text{on } (-\infty,-3) \cup (6,\infty),$
$< 0\, \text{on } (-3,6).$
Now we combine signs to solve
$$\frac{\text{numerator}}{\text{denominator}} \ge 0.$$
We need either numerator $\ge 0$ and denominator $>0$, or numerator $\le 0$ and denominator $<0$.
From above: Numerator $\ge 0$ on $(-\infty,-2] \cup [-1,\infty)$, denominator $>0$ on $(-\infty,-3) \cup (6,\infty)$.
Intersections:
$$(-\infty,-2] \cap (-\infty,-3) = (-\infty,-3),$$
$$[-1,\infty) \cap (6,\infty) = (6,\infty).$$
So this case gives $(-\infty,-3) \cup (6,\infty)$.
Numerator $\le 0$ on $[-2,-1]$, denominator $<0$ on $(-3,6)$.
Intersection: $[-2,-1]$ (the denominator is negative and nonzero there).
Including the points where the numerator is zero ($x=-2,-1$) and excluding the points where the denominator is zero ($x=-3,6$), the full solution is
$$x \in (-\infty,-3)\ \cup\ [-2,-1]\ \cup\ (6,\infty).$$
Traditional factor–sign method
Factor numerator and denominator:
$$x^2 + 3x + 2 = (x+1)(x+2),\qquad x^2 - 3x - 18 = (x-6)(x+3).$$
The critical points (zeros and poles) are
$$x = -2,\,-1,\,-3,\,6.$$
A standard sign analysis on the intervals
$$(-\infty,-3),\ (-3,-2),\ (-2,-1),\ (-1,6),\ (6,\infty)$$
shows that the fraction is:
positive on $(-\infty,-3)$ and $(6,\infty)$,
negative on $(-3,-2)$ and $(-1,6)$,
positive again on $(-2,-1)$.
Because we have $\ge 0$, we include the zeros of the numerator (at $x=-2$ and $x=-1$) and exclude the zeros of the denominator (at $x=-3$ and $x=6$).
Thus the traditional factor–sign method gives the same solution set:
$$x \in (-\infty,-3)\ \cup\ [-2,-1]\ \cup\ (6,\infty).$$
