Polynomial, fractional and cubic inequalities

Text

Solve the following inequalities, using sign tables where needed:

• (a) $x^3+3+x>5x^2$
• (b) $(x-2)(x+3) \le (x-2)(2x+5)$
• (c) $-\dfrac{1}{x-2}+x-2\le \dfrac{3}{2}$
• (d) $\dfrac{(-x+3)(4-3x)}{x(x-1)^2}>0$
• (e) $\dfrac{x^3+x^2+x}{x^2-x+1}\ge 0$

Solution

(a) $x^3-5x^2+x+3>0$. One root is $x=1$; dividing gives $(x-1)(x^2-4x-3)>0$, with $x^2-4x-3=0 \Rightarrow x=2\pm\sqrt7$. Ordered roots $2-\sqrt7\approx-0{,}65$, $1$, $2+\sqrt7\approx4{,}65$. Studying the sign of the product: $2-\sqrt7<x<1 \ \vee\ x>2+\sqrt7$.

(b) $(x-2)(x+3)-(x-2)(2x+5)\le 0 \implies (x-2)(-x-2)\le 0 \implies (x-2)(x+2)\ge 0 \implies x\le -2 \ \vee\ x\ge 2$.

(c) Setting $t=x-2$: $-\dfrac1t+t\le\dfrac32 \implies \dfrac{2t^2-3t-2}{2t}\le 0 \implies \dfrac{(2t+1)(t-2)}{2t}\le 0$, with solution $t\le -\tfrac12 \ \vee\ 0<t\le 2$. Returning to $x=t+2$: $x\le \tfrac32 \ \vee\ 2<x\le 4$.

(d) $\dfrac{(-x+3)(4-3x)}{x(x-1)^2}>0$. Since $(x-1)^2>0$ (with $x\neq 1$), the sign is that of $\dfrac{(3-x)(4-3x)}{x}$. Breakpoints $0$ (excluded), $\tfrac43$, $3$, and $x\neq 1$: solution $0<x<1 \ \vee\ 1<x<\tfrac43 \ \vee\ x>3$.

(e) $\dfrac{x^3+x^2+x}{x^2-x+1}\ge 0$. Numerator $x(x^2+x+1)$ and denominator $x^2-x+1$: both trinomials have $\Delta<0$ and are positive, so the sign is that of $x$. Solution $x\ge 0$.

$\boxed{\begin{aligned} &\text{(a) } 2-\sqrt7<x<1 \vee x>2+\sqrt7; \quad \text{(b) } x\le -2 \vee x\ge 2;\\ &\text{(c) } x\le \tfrac32 \vee 2<x\le 4; \quad \text{(d) } 0<x<1 \vee 1<x<\tfrac43 \vee x>3; \quad \text{(e) } x\ge 0\end{aligned}}$