Fractional inequalities (II)

Text

Solve:

• (a) $\dfrac{3-x}{1+x^2}\le 0$
• (b) $\dfrac{(25-x^2)(1-x^2)}{x^2+4}\ge 0$
• (c) $\dfrac{2}{3-x}-\dfrac{4x}{(1-x)^2}< \dfrac{x}{(1-x)(3-x)}$

Solution

(a) The denominator $1+x^2>0$ always, so the inequality is equivalent to $3-x\le 0 \iff x\ge 3$.

(b) Since $x^2+4>0$, there remains $(25-x^2)(1-x^2)\ge 0 \iff (5-x)(5+x)(1-x)(1+x)\ge 0$. Equivalently $(x^2-25)(x^2-1)\ge 0$. Critical points: $-5,\ -1,\ 1,\ 5$.

	$x<-5$	$-5<x<-1$	$-1<x<1$	$1<x<5$	$x>5$
$x^2-25$	$+$	$-$	$-$	$-$	$+$
$x^2-1$	$+$	$+$	$-$	$+$	$+$
product	$+$	$-$	$+$	$-$	$+$

Solution: $x\le -5\ \vee\ -1\le x\le 1\ \vee\ x\ge 5$.

(c) Domain: $x\neq 1,\ x\neq 3$. Over the common denominator $(1-x)^2(3-x)$: $\frac{2(1-x)^2-4x(3-x)-x(1-x)}{(1-x)^2(3-x)}< 0.$ The numerator is $(2-4x+2x^2)+(-12x+4x^2)+(-x+x^2)=7x^2-17x+2$, hence $\dfrac{7x^2-17x+2}{(1-x)^2(3-x)}<0$. Since $(1-x)^2>0$ (with $x\neq 1$), the sign depends on $\dfrac{7x^2-17x+2}{3-x}<0$. Roots of the numerator: $x=\dfrac{17\pm\sqrt{233}}{14}$ (numerically $\approx 0.124$ and $\approx 2.305$). Ordered critical points: $\dfrac{17-\sqrt{233}}{14},\ \dfrac{17+\sqrt{233}}{14},\ 3$.

	$x<\frac{17-\sqrt{233}}{14}$	$\frac{17-\sqrt{233}}{14}<x<\frac{17+\sqrt{233}}{14}$	$\frac{17+\sqrt{233}}{14}<x<3$	$x>3$
$7x^2-17x+2$	$+$	$-$	$+$	$+$
$3-x$	$+$	$+$	$+$	$-$
ratio	$+$	$-$	$+$	$-$

The ratio is $<0$ (strict inequality) for $\dfrac{17-\sqrt{233}}{14}<x<\dfrac{17+\sqrt{233}}{14}$ (with $x\neq 1$) $\ \vee\ x>3$.

$\boxed{\begin{aligned} &\text{(a) } x\ge 3; \quad \text{(b) } x\le -5 \vee -1\le x\le 1 \vee x\ge 5;\\ &\text{(c) } \tfrac{17-\sqrt{233}}{14}<x<\tfrac{17+\sqrt{233}}{14}\ (x\neq 1)\ \vee\ x>3\end{aligned}}$