Quadratic Inequalities: Difficult Problems with Solutions

Solution:
[tex]x^2-\left(1+\sqrt{3}\right)x+\sqrt{3}\lt 0[/tex]
The sum of the roots is [tex]1+\sqrt{3}[/tex] and their product is [tex]\sqrt{3}[/tex], so by Vieta's formulas the roots are [tex]1[/tex] and [tex]\sqrt{3}[/tex]:
[tex](x-1)\left(x-\sqrt{3}\right)\lt 0 \Leftrightarrow 1\lt x\lt \sqrt{3}[/tex]
Answer: [tex]x\in\left(1, \sqrt{3}\right)[/tex].

Solution:
Multiply both sides by [tex]10[/tex] (the least common denominator):
[tex]2(x+3)^2-5x(2x-3)\leq 2(3x-1)^2-10[/tex]
[tex]2x^2+12x+18-10x^2+15x\leq 18x^2-12x+2-10[/tex]
[tex]0\leq 26x^2-39x-26 \Leftrightarrow 2x^2-3x-2\geq 0 \Leftrightarrow (2x+1)(x-2)\geq 0[/tex]
Answer: [tex]x\in\left(-\infty, -\frac{1}{2}\right]\cup[2, +\infty)[/tex].

Solution:
[tex]4x^2+4x^2-4\sqrt{3}\,x+3-\left(x^2+\sqrt{3}\,x-6\right)\lt 63[/tex]
[tex]7x^2-5\sqrt{3}\,x-54\lt 0[/tex]
[tex]D=75+4\times 7\times 54=1587=3\times 23^2[/tex], [tex]\sqrt{D}=23\sqrt{3}[/tex]
[tex]x_{1,2}=\frac{5\sqrt{3}\pm 23\sqrt{3}}{14}[/tex]: [tex]x_1=-\frac{9\sqrt{3}}{7}[/tex], [tex]x_2=2\sqrt{3}[/tex]
Answer: [tex]x\in\left(-\frac{9\sqrt{3}}{7}, 2\sqrt{3}\right)[/tex].

Solution:
A double inequality is a system of two inequalities:
[tex]x^2+x\gt 0 \Leftrightarrow x(x+1)\gt 0 \Leftrightarrow x\lt -1[/tex] or [tex]x\gt 0[/tex]
[tex]x^2-2x-1\lt 0 \Leftrightarrow 1-\sqrt{2}\lt x\lt 1+\sqrt{2}[/tex]
Since [tex]1-\sqrt{2}\approx -0. 41\gt -1[/tex], the intersection is [tex]x\in\left(0, 1+\sqrt{2}\right)[/tex].

Solution:
[tex]x^2-1\geq 0 \Leftrightarrow x\leq -1[/tex] or [tex]x\geq 1[/tex]
[tex]x^3+x^2-2x=x(x+2)(x-1)\leq 0 \Leftrightarrow x\in(-\infty, -2]\cup[0, 1][/tex]
The solution of the system is the intersection (common part) of the two sets:
Answer: [tex]x\in(-\infty, -2]\cup\{1\}[/tex].
The number [tex]1[/tex] belongs to both sets, so it is an isolated solution.

Solution:
Let [tex]t=x^2\geq 0[/tex]: [tex]t^2-10t+16=(t-2)(t-8)\geq 0 \Leftrightarrow t\leq 2[/tex] or [tex]t\geq 8[/tex].
[tex]x^2\leq 2 \Leftrightarrow -\sqrt{2}\leq x\leq\sqrt{2}[/tex]
[tex]x^2\geq 8 \Leftrightarrow x\leq -2\sqrt{2}[/tex] or [tex]x\geq 2\sqrt{2}[/tex]
Answer: [tex]x\in\left(-\infty, -2\sqrt{2}\right]\cup\left[-\sqrt{2}, \sqrt{2}\right]\cup\left[2\sqrt{2}, +\infty\right)[/tex].

Solution:
Let [tex]y=x^2+3x[/tex]: [tex](y+1)(y-3)\geq 5 \Leftrightarrow y^2-2y-8\geq 0 \Leftrightarrow (y-4)(y+2)\geq 0[/tex].
[tex]y\leq -2:\ x^2+3x+2\leq 0 \Leftrightarrow (x+1)(x+2)\leq 0 \Leftrightarrow -2\leq x\leq -1[/tex]
[tex]y\geq 4:\ x^2+3x-4\geq 0 \Leftrightarrow (x+4)(x-1)\geq 0 \Leftrightarrow x\leq -4[/tex] or [tex]x\geq 1[/tex]
Answer: [tex]x\in(-\infty, -4]\cup[-2, -1]\cup[1, +\infty)[/tex].

Solution:
Consider the left side as a quadratic trinomial in [tex]a[/tex]:
[tex]D=b^2+8b^2=9b^2[/tex], [tex]a_{1,2}=\frac{b\pm 3b}{4}[/tex]: [tex]a_1=-\frac{b}{2}[/tex], [tex]a_2=b[/tex].
[tex]2a^2-ab-b^2=(2a+b)(a-b)\lt 0[/tex]
Since [tex]b\gt 0[/tex], we have [tex]-\frac{b}{2}\lt b[/tex], so [tex]-\frac{b}{2}\lt a\lt b[/tex].

Solution:
The leading coefficient is [tex]1\gt 0[/tex], so the trinomial is positive for every [tex]x[/tex] exactly when it has no real roots, i.e. [tex]D\lt 0[/tex]:
[tex]\frac{D}{4}=(m+2)^2-(m+8)=m^2+3m-4\lt 0[/tex]
[tex](m+4)(m-1)\lt 0 \Leftrightarrow -4\lt m\lt 1[/tex]
Answer: [tex]m\in(-4, 1)[/tex].

Solution:
The abscissas of the common points are the roots of the equation
[tex]4x^2-3ax+2=x^2-1 \Leftrightarrow 3x^2-3ax+3=0 \Leftrightarrow x^2-ax+1=0[/tex]
There are no common points when this equation has no real roots: [tex]D=a^2-4\lt 0 \Leftrightarrow -2\lt a\lt 2[/tex].
Answer: [tex]a\in(-2, 2)[/tex].

Solution:
[tex]x^2-(m+1)x+m=(x-1)(x-m)[/tex]
If [tex]m=1[/tex], the inequality [tex](x-1)^2\lt 0[/tex] has no solutions. If [tex]m\lt 1[/tex], the solutions are [tex](m, 1)[/tex], which is not inside [tex](1, 3)[/tex]. If [tex]m\gt 1[/tex], the solutions are [tex](1, m)[/tex], and this interval lies in [tex](1, 3)[/tex] exactly when [tex]m\leq 3[/tex].
Answer: [tex]m\in(1, 3][/tex].