A quadratic inequality is an inequality of the form $ax^2+bx+c>0$ (or with $\ge$, $<$, $\le$), where $a\ne 0$. The key to solving it is knowing the sign of the trinomial $y=ax^2+bx+c$. This sign is completely determined by two things: the discriminant $D=b^2-4ac$ and the sign of the leading coefficient $a$ (the direction of the parabola's branches). If $D>0$, the parabola crosses the $x$-axis at two points $x_1<x_2$ and the trinomial changes sign at them. If $D=0$, the parabola touches the axis at one point $x_0=-\frac{b}{2a}$ and does not change sign. If $D<0$, the parabola does not meet the axis at all, so the trinomial keeps the sign of $a$ for every real $x$.
Let $D=b^2-4ac$ and, when $D>0$, let $x_1<x_2$ be the roots.$$D>0,\ a>0:\quad ax^2+bx+c>0 \iff x<x_1 \ \text{or}\ x>x_2,\qquad ax^2+bx+c<0 \iff x_1<x<x_2$$$$D>0,\ a<0:\quad ax^2+bx+c>0 \iff x_1<x<x_2,\qquad ax^2+bx+c<0 \iff x<x_1 \ \text{or}\ x>x_2$$$$D=0:\quad ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2 \ \text{has the sign of}\ a \ \text{for all}\ x\ne -\frac{b}{2a},\ \text{and equals}\ 0 \ \text{at}\ x=-\frac{b}{2a}$$$$D<0,\ a>0:\quad ax^2+bx+c>0 \ \text{for all}\ x\in\mathbb{R};\qquad D<0,\ a<0:\quad ax^2+bx+c<0 \ \text{for all}\ x\in\mathbb{R}$$Consequence: for $D<0$, the inequality matching the sign of $a$ has solution $\mathbb{R}$, and the opposite inequality has no solutions ($\varnothing$).
Algorithm: 1) bring the inequality to the form $ax^2+bx+c$ compared with $0$; 2) compute $D$; 3) if $D>0$, find the roots and use the parabola picture; 4) if $D\le 0$, answer immediately from the sign of $a$. In YÖS problems the case $D<0$ appears very often. Example: $x^2-x+1>0$ has $D=1-4=-3<0$ and $a=1>0$, so the trinomial is positive everywhere and the answer is $x\in\mathbb{R}$. Example: $x^2-x+6<0$ has $D=1-24=-23<0$ and $a>0$, so the trinomial is always positive and can never be negative: the solution set is empty, $\varnothing$. Do not waste time trying to factor such trinomials — check $D$ first.
The interval method solves any inequality of the form $\frac{P(x)}{Q(x)}$ compared with $0$, where $P$ and $Q$ are products of factors. Steps: 1) move everything to one side so that the right side is $0$; 2) factor the numerator and the denominator; 3) find all zeros of the numerator and of the denominator — these are the critical points; 4) mark them on the number line: zeros of the numerator are filled dots for $\ge,\le$ and empty for $>,<$; zeros of the denominator are ALWAYS empty dots; 5) determine the sign on the rightmost interval (take a large test value of $x$); 6) moving left, the sign alternates when passing a point of odd multiplicity and stays the same at a point of even multiplicity; 7) select the intervals with the required sign.
If a factor $(x-a)^n$ has an odd exponent $n$, the expression changes sign at $x=a$; if $n$ is even, the sign does not change (the point only gives the value $0$ if it is a zero of the numerator). For huge exponents only the parity matters: $(x-1)^{1938}$ behaves like $(x-1)^2$, and $(2-x)^{1881}$ behaves like $(2-x)$. Also watch for factors with constant sign: $x^2+4>0$ and $x^2+x+1>0$ for all $x$ (since $D<0$), while $-x^2-4=-(x^2+4)<0$ for all $x$. Such a factor can be removed from the inequality; if it is negative, flip the inequality sign. Useful factorizations: $x^3-1=(x-1)(x^2+x+1)$, where the second factor is always positive, and $x^2-9=(x-3)(x+3)$. Finally, a pure square such as $\frac{(x-2)^2}{(x+3)^2}$ is $\ge 0$ automatically: the inequality $\ge 0$ holds for every $x$ in the domain, i.e. for all $x\ne -3$.
The most common mistake is multiplying both sides by the denominator. The denominator contains $x$, its sign is unknown, so multiplication may or may not flip the inequality — NEVER do it. Correct method: move everything to one side and combine into a single fraction. Example: $\frac{x}{x+2}\le\frac{1}{x}$ must NOT be turned into $x^2\le x+2$; instead write $\frac{x}{x+2}-\frac{1}{x}\le 0$, i.e. $\frac{x^2-(x+2)}{x(x+2)}\le 0$, i.e. $\frac{(x-2)(x+1)}{x(x+2)}\le 0$, and apply the interval method. Boundary points rule: zeros of the numerator are included when the sign is $\ge$ or $\le$; zeros of the denominator (here $x=0$ and $x=-2$) are excluded ALWAYS, even with $\ge$ or $\le$, because division by zero is undefined.
A double inequality $A\le f(x)\le B$ is a system of two inequalities that must hold simultaneously: $f(x)\ge A$ AND $f(x)\le B$. Solve each one separately and intersect the answers. Scheme for $-2\le x^2-2x-5\le 10$: the left part gives $x^2-2x-3\ge 0$, i.e. $(x-3)(x+1)\ge 0$, so $x\le -1$ or $x\ge 3$; the right part gives $x^2-2x-15\le 0$, i.e. $(x-5)(x+3)\le 0$, so $-3\le x\le 5$. The intersection of the two sets is the final answer: $[-3,-1]\cup[3,5]$. Never chain-solve a double inequality with a quadratic middle as if it were linear.
For $\sqrt{f(x)}<a$ with $a>0$, two conditions must hold together: the domain condition $f(x)\ge 0$ (the expression under the root must be non-negative) and, after squaring both non-negative sides, $f(x)<a^2$. So $\sqrt{f(x)}<a \iff 0\le f(x)<a^2$. Example: $\sqrt{x^2+3x-4}<6$. Domain: $x^2+3x-4\ge 0$, i.e. $(x+4)(x-1)\ge 0$, so $x\le -4$ or $x\ge 1$. Squaring: $x^2+3x-4<36$, i.e. $x^2+3x-40<0$, i.e. $(x+8)(x-5)<0$, so $-8<x<5$. Intersecting both conditions: $x\in(-8,-4]\cup[1,5)$. Note: for $\sqrt{f(x)}>a$ with $a\ge 0$ the answer is simply $f(x)>a^2$ (the domain condition is then automatic).
Standard YÖS formulations. 1) The statement that $ax^2+bx+c>0$ for all $x\in\mathbb{R}$ is equivalent to $a>0$ and $D<0$ (an upward parabola entirely above the axis). Similarly, negativity for all $x$ is equivalent to $a<0$ and $D<0$. 2) The equation has two distinct real roots $\iff D>0$; it has no real roots $\iff D<0$; it has a repeated root $\iff D=0$. Example: find all $m$ such that $x^2-(m+2)x+7>0$ for every $x$. Here $a=1>0$, so we need $D<0$: $(m+2)^2-4\cdot 7<0$, i.e. $(m+2)^2<28$, i.e. $|m+2|<2\sqrt{7}$, giving $-2-2\sqrt{7}<m<-2+2\sqrt{7}$. If instead the problem asks for two distinct roots of $x^2-(m+2)x+7=0$, solve $(m+2)^2>28$: $m<-2-2\sqrt{7}$ or $m>-2+2\sqrt{7}$.
Many YÖS problems ask how many integers $x\in\mathbb{Z}$ satisfy the inequality, or ask for the sum of all integer solutions. Procedure: solve the inequality completely, then carefully list the integers inside each interval of the answer. Pay attention to the endpoints: an endpoint enters the list only if the interval is closed at it (filled dot); zeros of a denominator never enter. Example: if the solution set is $[-3,-1]\cup[3,5]$, the integers are $-3,-2,-1,3,4,5$ and their sum is $(-3-2-1)+(3+4+5)=-6+12=6$. Speed trick: on a symmetric run of consecutive integers from $-n$ to $n$ the sum is $0$, so only the unbalanced tail contributes.
Problem. Solve: a) $x^2-x+1>0$; b) $x^2-x+6<0$.Solution a). Step 1: compute the discriminant: $D=(-1)^2-4\cdot 1\cdot 1=-3<0$. Step 2: since $a=1>0$ and $D<0$, the parabola lies entirely above the $x$-axis, so $x^2-x+1>0$ for every real $x$. Answer: $x\in\mathbb{R}$, i.e. $(-\infty,+\infty)$.Solution b). Step 1: $D=(-1)^2-4\cdot 1\cdot 6=1-24=-23<0$. Step 2: $a=1>0$ and $D<0$ mean the trinomial is strictly positive for all $x$, hence it is never negative. Answer: no solutions, $\varnothing$.
Problem. Solve $\frac{x}{x+2}\le\frac{1}{x}$.Step 1. Do NOT multiply by $x(x+2)$ — its sign is unknown. Move everything to the left: $\frac{x}{x+2}-\frac{1}{x}\le 0$.Step 2. Common denominator: $\frac{x\cdot x-(x+2)}{x(x+2)}\le 0$, i.e. $\frac{x^2-x-2}{x(x+2)}\le 0$.Step 3. Factor the numerator: $x^2-x-2=(x-2)(x+1)$, so $\frac{(x-2)(x+1)}{x(x+2)}\le 0$.Step 4. Critical points: numerator zeros $x=-1$ and $x=2$ (filled, since the sign is $\le$); denominator zeros $x=-2$ and $x=0$ (always excluded). All multiplicities are odd, so the sign alternates: for $x>2$ the fraction is positive, on $(0,2)$ negative, on $(-1,0)$ positive, on $(-2,-1)$ negative, for $x<-2$ positive.Step 5. We need $\le 0$: the negative intervals plus the numerator zeros. Answer: $x\in(-2,-1]\cup(0,2]$.
Problem. Solve $\frac{(2-x)^{1881}(x-1)^{1938}}{(x+1)^{55}(x^2-9)}\ge 0$.Step 1. Only the parity of the exponents matters: $1881$ and $55$ are odd (sign changes), $1938$ is even (no sign change). Factor: $x^2-9=(x-3)(x+3)$.Step 2. Critical points: numerator zeros $x=2$ (odd multiplicity) and $x=1$ (even) — candidates for inclusion since the sign is $\ge$; denominator zeros $x=-1$, $x=3$, $x=-3$ (all odd multiplicity) — always excluded.Step 3. Sign on the rightmost interval $x>3$: $(2-x)^{1881}<0$, $(x-1)^{1938}>0$, $(x+1)^{55}>0$, $x^2-9>0$, so the whole fraction is negative.Step 4. Moving left: at $x=3$ (odd) the sign flips to positive on $(2,3)$; at $x=2$ (odd) it flips to negative on $(1,2)$; at $x=1$ (even) it does NOT flip, staying negative on $(-1,1)$; at $x=-1$ (odd) it flips to positive on $(-3,-1)$; at $x=-3$ (odd) it flips to negative for $x<-3$.Step 5. We need $\ge 0$: the positive intervals $(-3,-1)$ and $(2,3)$, plus the numerator zeros where the fraction is defined: $x=2$ (included) and the isolated point $x=1$ (the fraction equals $0$ there). The points $-3,-1,3$ are excluded. Answer: $x\in(-3,-1)\cup\{1\}\cup[2,3)$.
Problem. Solve $-2\le x^2-2x-5\le 10$ and find the sum of all integer solutions.Step 1. Split into a system: $x^2-2x-5\ge -2$ and $x^2-2x-5\le 10$.Step 2. Left inequality: $x^2-2x-3\ge 0$, i.e. $(x-3)(x+1)\ge 0$, so $x\le -1$ or $x\ge 3$.Step 3. Right inequality: $x^2-2x-15\le 0$, i.e. $(x-5)(x+3)\le 0$, so $-3\le x\le 5$.Step 4. Intersect: $\big((-\infty,-1]\cup[3,+\infty)\big)\cap[-3,5]=[-3,-1]\cup[3,5]$.Step 5. Integer solutions: $-3,-2,-1,3,4,5$. Their sum: $(-3)+(-2)+(-1)+3+4+5=6$. Answer: $x\in[-3,-1]\cup[3,5]$; the sum of the integer solutions is $6$.
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