Second-degree and fractional inequalities (variant)

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Solve the following inequalities:

• (a) $x^2-x-1\ge 0$
• (b) $x^2+6x+9 \le 0$
• (c) $x^2+2x-3\le 0$
• (d) $\dfrac{-x^2+6x-9}{x^2+2x-3}>0$
• (e) $\dfrac{3}{1-z}+3-3z\le 9$

Solution

(a) $x^2-x-1=0 \Rightarrow x=\dfrac{1\pm\sqrt5}{2}$; upward concavity, so $\ge 0$ outside the roots: $x\le \dfrac{1-\sqrt5}{2} \ \vee\ x\ge \dfrac{1+\sqrt5}{2}$.

(b) $x^2+6x+9=(x+3)^2\le 0$: only $x=-3$.

(c) $x^2+2x-3=(x+3)(x-1)\le 0 \implies -3\le x\le 1$.

(d) $\dfrac{-x^2+6x-9}{x^2+2x-3}=\dfrac{-(x-3)^2}{(x+3)(x-1)}$. The numerator is $\le 0$ (zero only at $x=3$), so the fraction is $>0$ only if the denominator is $<0$: $(x+3)(x-1)<0 \implies -3<x<1$ (where $x=3$ does not occur).

(e) $\dfrac{3}{1-z}+3-3z-9\le 0 \implies \dfrac{3(z^2+z-1)}{1-z}\le 0 \implies \dfrac{z^2+z-1}{1-z}\le 0$. Numerator roots $z=\dfrac{-1\pm\sqrt5}{2}$, denominator zero at $z=1$. Sign study: $\dfrac{-1-\sqrt5}{2}\le z\le \dfrac{-1+\sqrt5}{2} \ \vee\ z>1$.

$\boxed{\begin{aligned} &\text{(a) } x\le \tfrac{1-\sqrt5}{2} \vee x\ge \tfrac{1+\sqrt5}{2}; \quad \text{(b) } x=-3; \quad \text{(c) } -3\le x\le 1;\\ &\text{(d) } -3<x<1; \quad \text{(e) } \tfrac{-1-\sqrt5}{2}\le z\le \tfrac{-1+\sqrt5}{2} \vee z>1\end{aligned}}$