Irrational Inequalities: Difficult Problems with Solutions

Solution:
Domain: [tex](2x+3)(x-3)\ge 0 \iff x\le -\dfrac{3}{2}[/tex] or [tex]x\ge 3[/tex].
[tex]x\lt 3[/tex]: the right side is negative and the inequality holds for [tex]x\le -\dfrac{3}{2}[/tex].
[tex]x\ge 3:\ 2x^2-3x-9\gt x^2-6x+9 \iff x^2+3x-18\gt 0 \iff (x+6)(x-3)\gt 0 \Rightarrow x\gt 3[/tex]
Answer: [tex]x\in\left(-\infty, -\dfrac{3}{2}\right]\cup(3, +\infty)[/tex]

Solution:
[tex]\begin{cases}(2x-1)(x-3)\ge 0\\ 3-x\ge 0\\ 2x^2-7x+3\le (3-x)^2\end{cases}[/tex]
[tex]2x^2-7x+3\le 9-6x+x^2 \iff x^2-x-6\le 0 \iff -2\le x\le 3[/tex]
[tex](2x-1)(x-3)\ge 0 \iff x\le \dfrac{1}{2}[/tex] or [tex]x\ge 3[/tex]
Answer: [tex]x\in\left[-2, \dfrac{1}{2}\right]\cup\{3\}[/tex]

Solution:
Domain: [tex]x\gt -18, x\ne 2[/tex].
[tex]x\gt 2:\ \dfrac{1}{2-x}\lt 0\lt \dfrac{1}{\sqrt{x+18}}\ \checkmark[/tex]
[tex]-18\lt x\lt 2:\ \sqrt{x+18}\lt 2-x \iff x+18\lt 4-4x+x^2 \iff x^2-5x-14\gt 0 \iff (x-7)(x+2)\gt 0 \Rightarrow x\lt -2[/tex]
Answer: [tex]x\in(-18, -2)\cup(2, +\infty)[/tex]

Solution:
Domain: [tex]12x-3x^2\ge 0 \iff 0\le x\le 4 \Rightarrow 6-x\gt 0[/tex].
Squaring: [tex]12x-3x^2\ge 36-12x+x^2 \iff 4x^2-24x+36\le 0 \iff 4(x-3)^2\le 0 \Rightarrow x=3[/tex]
Answer: [tex]x=3[/tex]

Solution:
[tex]x^2-3x+5\gt 0[/tex] for every [tex]x\ (D=-11\lt 0)[/tex].
Substitute [tex]t=\sqrt{x^2-3x+5}\gt 0, x^2-3x=t^2-5[/tex]: [tex]7\lt t^2-5+t \iff t^2+t-12\gt 0 \iff (t+4)(t-3)\gt 0 \Rightarrow t\gt 3[/tex]
[tex]x^2-3x+5\gt 9 \iff (x-4)(x+1)\gt 0[/tex]
Answer: [tex]x\in(-\infty, -1)\cup(4, +\infty)[/tex]

Solution:
Substitute [tex]t=\sqrt{x}\ge 0, t\ne 5[/tex]: [tex]\dfrac{4-(1+t)(5-t)}{5-t}=\dfrac{t^2-4t-1}{5-t}\ge 0[/tex]
[tex]t^2-4t-1=0 \Rightarrow t=2\pm\sqrt{5}, t\ge 0 \Rightarrow 2+\sqrt{5}\le t\lt 5[/tex]
[tex]x=t^2\in\left[\left(2+\sqrt{5}\right)^2, 25\right)=\left[9+4\sqrt{5}, 25\right)[/tex]

Solution:
Domain: [tex]-2\le x\le 2, x^2+2x\ge 0 \Rightarrow x\in\{-2\}\cup[0, 2][/tex].
[tex]x=-2:\ 0+0\lt 2\ \checkmark[/tex]
[tex]x\in[0, 2]:\ \left(\sqrt{4-x^2}+\sqrt{x^2+2x}\right)^2=4+2x+2\sqrt{\left(4-x^2\right)\left(x^2+2x\right)}\ge 4[/tex] — no solution.
Answer: [tex]x=-2[/tex]

Solution:
Domain: [tex]-2\le x\le 3, x\ne \dfrac{5}{2}[/tex].
[tex]x=-2[/tex] or [tex]x=3:\ 0\le 0\ \checkmark[/tex]
[tex]-2\lt x\lt 3:\ \dfrac{1}{x+4}\le \dfrac{1}{2x-5}, x+4\gt 0[/tex]
[tex]x\lt \dfrac{5}{2}:\ \dfrac{1}{2x-5}\lt 0\lt \dfrac{1}{x+4}[/tex] — no solution.
[tex]\dfrac{5}{2}\lt x\lt 3:\ 2x-5\le x+4 \iff x\le 9\ \checkmark[/tex]
Answer: [tex]x\in\{-2\}\cup\left(\dfrac{5}{2}, 3\right][/tex]

Solution:
[tex]\sqrt{4x^2+12x+9}=|2x+3|, 2x^2-8x+10=2(x-2)^2+2\gt 0[/tex]
[tex]x\ge -\dfrac{3}{2}:\ x+1\gt \sqrt{2x^2-8x+10}, x\gt -1:\ x^2+2x+1\gt 2x^2-8x+10 \iff x^2-10x+9\lt 0 \iff 1\lt x\lt 9[/tex]
[tex]x\lt -\dfrac{3}{2}:\ 5x+7\gt \sqrt{2x^2-8x+10}\gt 0 \Rightarrow x\gt -\dfrac{7}{5}[/tex] — impossible.
The integer solutions are [tex]2, 3, \ldots, 8[/tex].
Answer: [tex]7[/tex]

Solution:
Domain: [tex]x\ne 1, x\ne 3[/tex].
The cube root is odd and increasing, so [tex]\sqrt[3]{u}+\sqrt[3]{v}\ge 0 \iff u\ge -v \iff u+v\ge 0[/tex].
[tex]u+v=\dfrac{(x+1)(x-1)+(x+3)(x-3)}{(x-3)(x-1)}=\dfrac{2\left(x^2-5\right)}{(x-1)(x-3)}\ge 0[/tex]
Answer: [tex]x\in\left(-\infty, -\sqrt{5}\right]\cup\left(1, \sqrt{5}\right]\cup(3, +\infty)[/tex]

Solution:
Domain: [tex]x\ge -\dfrac{10}{9}, \sqrt{9x+10}\ne -3x \iff x\ne -\dfrac{2}{3}[/tex].
[tex]9x+10-9x^2=\left(\sqrt{9x+10}-3x\right)\left(\sqrt{9x+10}+3x\right) \Rightarrow \sqrt{9x+10}-3x\ge 0[/tex]
[tex]x\lt 0\ \checkmark[/tex]; [tex]x\ge 0:\ 9x+10\ge 9x^2 \iff (3x+2)(3x-5)\le 0 \Rightarrow 0\le x\le \dfrac{5}{3}[/tex]
Answer: [tex]x\in\left[-\dfrac{10}{9}, -\dfrac{2}{3}\right)\cup\left(-\dfrac{2}{3}, \dfrac{5}{3}\right][/tex]