Quadratic and rational inequalities

Quadratic inequalities. To solve $ax^2+bx+c \lessgtr 0$:

• Get zero on one side, then factor or find the roots $\alpha \le \beta$ of $ax^2+bx+c=0$.
• If $a>0$, the expression is negative between the roots and positive outside them; if $a<0$, the pattern reverses. This single rule replaces the need to memorise separate cases for $>$, $<$, $\ge$, $\le$.
• Strict inequalities ($>,<$) exclude the roots themselves; non-strict inequalities ($\ge,\le$) include them.

Rational inequalities. For $\dfrac{f(x)}{g(x)} \lessgtr a$, the single most dangerous move is cross-multiplying by $g(x)$ — its sign is usually unknown, so multiplying can silently flip the inequality for some values of $x$ and wrong for others. The safe route brings everything to one side first:

where: $f(x)$ and $g(x)$ are polynomials and $a$ is a constant (often $0$). Now the critical points are the zeros of the new numerator and the zeros of $g(x)$ — the denominator's zeros are always excluded from the final answer, since the expression is undefined there, regardless of the inequality symbol.

Repeated roots. If a critical point comes from a factor raised to an even power, such as $(x-2)^2$, the sign does not change as $x$ crosses it — the curve touches zero and bounces back. An odd-power factor (including a simple linear factor) does flip the sign. This is precisely why a test point in every region beats blind alternation.

Solving $ax^2+bx+c \lessgtr 0$ (with $a\ne0$ and real, distinct roots $\alpha<\beta$) by the wavy-curve method rests on writing the quadratic as a product of two linear factors and tracking how a product's sign responds as each factor individually changes sign while crossing a simple root; the goal here is to show algebraically why this alternation happens and why it produces the familiar "outside positive, inside negative" pattern for $a>0$.

Step 1. Factor the quadratic using its roots: $ax^2+bx+c = a(x-\alpha)(x-\beta)$, with $\alpha<\beta$. This is always possible once the roots are known, by the factor theorem, and it converts a quadratic sign question into a question about the sign of a product of simple linear pieces.

Step 2. Track each linear factor separately. For $x>\beta$ (to the right of both roots), both $(x-\alpha)>0$ and $(x-\beta)>0$, so their product is positive; multiplying by $a>0$ keeps it positive. For $\alpha<x<\beta$, $(x-\alpha)>0$ but $(x-\beta)<0$, so the product $(x-\alpha)(x-\beta)$ is negative, and $a$ times a negative number (with $a>0$) is negative. For $x<\alpha$, both factors are negative, so their product is positive again, and multiplying by $a>0$ keeps it positive.

Step 3. Read off the pattern: positive, negative, positive as $x$ moves from $-\infty$ through $\alpha$, between the roots, through $\beta$, to $+\infty$ — exactly one sign change at each simple root. This confirms the memorised rule "$a>0$: negative between the roots, positive outside" is not a coincidence but a direct consequence of a product of two linear factors changing sign exactly once at each of its zeros.

Step 4. The same reasoning extends to any product of distinct linear factors (three, four, or more critical points): each simple root flips the sign of exactly one factor, so the overall sign alternates once per critical point as you scan the number line — the general justification for the wavy-curve method used with rational expressions and higher-degree factorisations. Check: at $x=0$ (between $\alpha=-2,\beta=3$ for $a=1$), $(0-(-2))(0-3) = (2)(-3) = -6<0$ — negative, matching Step 3's middle-interval prediction.

Solve $\dfrac{2x-1}{x+3}\le1$, then check what happens if the inequality is tightened to $\dfrac{2x-1}{x+3}\le0$ using the same critical points.

Step 1. Bring everything to one side rather than cross-multiplying: $\dfrac{2x-1}{x+3}-1\le0 \implies \dfrac{2x-1-(x+3)}{x+3}\le0 \implies \dfrac{x-4}{x+3}\le0$.

Step 2. Identify critical points: the numerator is zero at $x=4$ (included, since the expression equals exactly $0$ there and the inequality is $\le$); the denominator is zero at $x=-3$ (always excluded, since the expression is undefined there).

Step 3. Test one point per region. For $x<-3$, say $x=-4$: $\dfrac{-4-4}{-4+3}=\dfrac{-8}{-1}=8>0$. For $-3<x<4$, say $x=0$: $\dfrac{0-4}{0+3}=\dfrac{-4}{3}<0$. For $x>4$, say $x=5$: $\dfrac{5-4}{5+3}=\dfrac{1}{8}>0$. The middle interval alone satisfies $\le0$.

Step 4. Assemble the answer: the middle region plus the included endpoint $x=4$ gives $-3<x\le4$, or $(-3,4]$ in interval notation. $x=-3$ is excluded even though it borders the solution, because the expression is undefined there.

Answer: $\boxed{(-3,4]}$.

Competitive stretch

Tighten the question to $\dfrac{2x-1}{x+3}\le0$ using the original expression (not the shifted one from Step 1). Critical points are the zeros of numerator $2x-1$ (i.e. $x=\tfrac12$, included) and denominator $x+3$ (i.e. $x=-3$, excluded). Testing regions exactly as before shows the expression is negative only for $-3<x<\tfrac12$, giving $\left(-3,\tfrac12\right]$ — a different answer from Worked example above, because subtracting $1$ before finding critical points genuinely changes which values of $x$ make the expression zero. This is a common way exam-setters build a second, related part to a question.

• Cross-multiplying $\dfrac{f(x)}{g(x)} \lessgtr a$ by $g(x)$ without knowing its sign — this can flip the inequality for only some values of $x$, silently corrupting the answer. Always bring everything to one side first.
• Including a denominator's zero in the final solution — the expression is undefined there and must be excluded no matter which inequality symbol is used.
• Assuming the sign always alternates at every critical point — a factor raised to an even power (a repeated root) touches zero and bounces back to the same sign instead of crossing.
• Forgetting the endpoint rule: strict inequalities exclude the critical points; non-strict inequalities include numerator-zero critical points but never denominator-zero ones.

Ask the tutor: If you get stuck deciding which critical points to include or exclude in the final interval, ask the AI tutor to walk through a rational inequality with you step by step, marking each point as included or excluded and explaining why.