4.3 Quadratic Equations and Inequalities from a Function Viewpoint
Key Takeaways
- Every quadratic equation $ax^2+bx+c=0$ corresponds to the zeros of $f(x)=ax^2+bx+c$, so equations, inequalities, and parameter problems are all questions about one graph.
- The discriminant $\Delta=b^2-4ac$ determines whether the parabola crosses the $x$-axis twice, touches it once, or never meets it.
- Quadratic inequalities are solved by locating the roots (or confirming none exist) and reading the sign of $f$ from a sign chart or parabola sketch, not by guessing.
- Parameter problems reason backward from a graphical condition — such as "always positive" or "two distinct roots" — to constraints on the unknown coefficient.
- Vertex form exposes the extremum value directly, and the vertex's $y$-coordinate equals $-\Delta/(4a)$, linking it back to the discriminant table.
Why View Equations and Inequalities as Functions?
Before Gaokao candidates can tackle quadratic optimization, inequality proofs, or discriminant-based parameter problems, they need one preparatory habit: read every quadratic equation or inequality as a statement about the graph of $f(x) = ax^2 + bx + c$ ($a \neq 0$). This function viewpoint converts algebra into geometry you can see, and it is the preparatory (预备知识) skill that underlies almost every later selective-compulsory topic.
The Zero-Function Connection
The real roots of $ax^2 + bx + c = 0$ are exactly the $x$-intercepts (zeros) of $f(x) = ax^2 + bx + c$. This single idea unifies three question types that look different on paper:
- Equation: "Solve $2x^2 - 5x + 2 = 0$" asks for the zeros of $f(x) = 2x^2 - 5x + 2$.
- Inequality: "Solve $2x^2 - 5x + 2 > 0$" asks where the parabola lies above the $x$-axis.
- Parameter: "For what values of $m$ does $mx^2 - 5x + 2 = 0$ have two distinct real roots?" asks when the parabola crosses the $x$-axis twice.
Worked Example 1. Solve $2x^2 - 5x + 2 = 0$ by finding the zeros of $f(x) = 2x^2 - 5x + 2$.
Factor: $2x^2 - 5x + 2 = (2x - 1)(x - 2)$. Setting each factor to zero gives $x = \frac{1}{2}$ and $x = 2$. So $f$ crosses the $x$-axis at $x = \frac{1}{2}$ and $x = 2$, and the solution set is ${\frac{1}{2}, 2}$.
Discriminant and the Number of Roots
The discriminant $\Delta = b^2 - 4ac$ tells you, before solving anything, how many times the parabola meets the $x$-axis:
| $\Delta$ value | Roots of $ax^2+bx+c=0$ | Graph behavior ($a>0$ shown) |
|---|---|---|
| $\Delta > 0$ | Two distinct real roots $x_1 \neq x_2$ | Parabola crosses the $x$-axis twice |
| $\Delta = 0$ | One repeated real root $x_1 = x_2 = -\frac{b}{2a}$ | Parabola touches the $x$-axis at its vertex |
| $\Delta < 0$ | No real roots | Parabola stays entirely above the $x$-axis |
Worked Example 2. For $f(x) = x^2 - 4x + 4$, $\Delta = 16 - 16 = 0$, so $f$ has a repeated root at $x = 2$ and the parabola is tangent to the $x$-axis at its vertex $(2, 0)$ — consistent with $x^2 - 4x + 4 = (x-2)^2$.
Solving Quadratic Inequalities via Sign Charts
Once you know the zeros (and hence the sign pattern) of $f$, quadratic inequalities become a graph-reading exercise:
- Rewrite the inequality so one side is $0$: $f(x) > 0$, $f(x) < 0$, $f(x) \ge 0$, or $f(x) \le 0$.
- Find the real roots of $f(x) = 0$ (or confirm there are none, via $\Delta$).
- Sketch the parabola's rough shape: opens up if $a > 0$, opens down if $a < 0$.
- Read the sign of $f$ directly from the picture: for $a > 0$, $f$ is negative between the roots and positive outside them; for $a < 0$, the reverse.
Worked Example 3. Solve $x^2 - x - 6 \le 0$.
Factor: $x^2 - x - 6 = (x-3)(x+2)$, roots $x = -2$ and $x = 3$. Since $a = 1 > 0$, the parabola opens upward and is $\le 0$ between its roots. Solution: $-2 \le x \le 3$, i.e. $[-2, 3]$.
Worked Example 4 (no real roots). Solve $x^2 + x + 1 > 0$.
$\Delta = 1 - 4 = -3 < 0$, so $f$ has no real zeros and never touches the $x$-axis. Since $a = 1 > 0$, the entire parabola sits above the $x$-axis, so the inequality holds for every real $x$: the solution set is $\mathbb{R}$.
Parameter Problems: Reasoning Backward from the Graph
Gaokao parameter questions run the function viewpoint in reverse: you are told a fact about the graph (crosses the axis twice, stays above it, has a root in an interval) and must find the parameter values that make it true.
Worked Example 5. Find all real $m$ such that $mx^2 - 2x + 1 > 0$ for every real $x$.
If $m = 0$, the expression becomes $-2x + 1$, which is linear and not always positive, so reject $m = 0$.
If $m \neq 0$, the parabola must stay entirely above the $x$-axis, which requires both $a > 0$ and $\Delta < 0$: $m > 0$ and $\Delta = 4 - 4m < 0 \implies m > 1$.
Combining the two conditions, the answer is $m > 1$.
Connection to Vertex Form
Completing the square rewrites $f(x) = ax^2 + bx + c$ as exposing the vertex at $\left(-\frac{b}{2a}, \frac{4ac-b^2}{4a}\right)$ directly. This form makes two facts obvious without extra computation: the maximum or minimum value of $f$ is $\frac{4ac-b^2}{4a}$ (a minimum when $a > 0$, a maximum when $a < 0$), and the vertex's $y$-coordinate equals exactly $-\frac{\Delta}{4a}$ — so the sign of $\Delta$ and the sign of $a$ together determine whether the parabola ever crosses the $x$-axis, matching the discriminant table above.
Moving fluently between factored form (roots visible), standard form ($\Delta$ visible), and vertex form (extremum visible) lets you pick whichever representation answers a given question fastest.
Gaokao Exam Trap
A common trap is treating a quadratic inequality like a linear one and "flipping the inequality when dividing by a coefficient" without checking the sign of $a$, or solving $ax^2+bx+c>0$ by writing "$x >$ root$_1$ and $x >$ root$_2$" as if both factors must be large. Always draw the parabola or build a short sign chart: the geometry, not a remembered verbal rule, is what keeps the answer set correct under time pressure.
What is the solution set of $x^2 - 5x + 6 < 0$?
For which values of $k$ (with $k \neq 0$) does $kx^2 - 4x + k = 0$ have no real roots?
The function $f(x) = -2x^2 + 8x - 3$ attains its maximum value at which point?