4.3 Quadratic Equations and Inequalities from a Function Viewpoint

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:

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:

1. Rewrite the inequality so one side is $0$: $f(x) > 0$, $f(x) < 0$, $f(x) \ge 0$, or $f(x) \le 0$.
2. Find the real roots of $f(x) = 0$ (or confirm there are none, via $\Delta$).
3. Sketch the parabola's rough shape: opens up if $a > 0$, opens down if $a < 0$.
4. 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

$$f(x) = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a},$$

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.