| English | 意味 |
|---|---|
| parabola/pəˈræbələ/ | 放物線 |
| quadratic/kwɒˈdrætɪk/ | 二次式 |
| completing the square/kəmˈpliːtɪŋ ðə skweə/ | 平方完成 |
| vertex/ˈvɜːteks/ | 頂点 |
| discriminant/dɪˈskrɪmɪnənt/ | 判別式 |
The path of a thrown ball
- Throw a ball and it traces a parabola 抛物线 — the same curve described by every quadratic 二次式 equation.
- Galileo proved this in 1638. Today, quadratics appear in physics, economics, engineering, and optimization.
- Mastering them means you can find the peak, the roots, and the shape of any parabola.
Completing the square 配方法
- A quadratic is $ax^2 + bx + c$ (with $a \neq 0$).
- Completing the square rewrites it as $a(x+p)^2 + q$.
- This reveals the vertex 顶点 (turning point) at $(-p,\ q)$.
- Example: $x^2 - 6x + 5 = (x-3)^2 - 9 + 5 = (x-3)^2 - 4$ → vertex $(3, -4)$.

Completing the square reveals the vertex directly: $y = (x-3)^2 - 4$ has its minimum at $(3, -4)$.
Worked example. Complete the square for $2x^2 + 8x + 3$: factor out $2$: $2(x^2 + 4x) + 3 = 2(x+2)^2 - 8 + 3 = 2(x+2)^2 - 5$. Vertex: $(-2, -5)$.
平方完成により、x² − 6x + 5 = (x − 3)² + q となる。最小値 q はいくつか?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
(x − 3)² − 9 + 5 = (x − 3)² − 4 より、q = −4(最小値)。
平方完成:2x² + 8x + 3 = 2(x + p)² + q。p はいくつか?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
2(x² + 4x) + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5 より、p = 2。
The discriminant 判别式
- The discriminant is $\Delta = b^2 - 4ac$. It counts the real roots:
- $b^2 - 4ac > 0$ → two distinct real roots,
- $b^2 - 4ac = 0$ → one repeated root,
- $b^2 - 4ac < 0$ → no real roots.
- The quadratic formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

The sign of the discriminant fixes how many times the parabola meets the x-axis: two roots, one, or none
The discriminant is under the square root. If $b^2 - 4ac < 0$, you'd need $\sqrt{\text{negative}}$, which has no real value — so the parabola never crosses the $x$-axis.

A suspension bridge: the main cable hangs in a parabola
放物線を探る
y = ax² + bx + c
a, b, c を変化させて曲線の形を変え、頂点、根、切片 が動く様子を見る。
x² + 4x + 1 = 0 について、判別式 b² − 4ac はいくつか?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
b² − 4ac = 4² − 4(1)(1) = 16 − 4 = 12 (> 0 なので、2つの異なる実根を持つ)。
b² − 4ac < 0 のとき、二次方程式には:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
負の判別式は放物線がx軸と交わらないことを意味し、実根を持たない。
方程式 x² + 2x + 5 = 0 は2つの異なる実根を持つ。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
判別式 = 2² − 4(1)(5) = 4 − 20 = −16 < 0 なので、実根はない。
Solving quadratics
- Factorising: fastest when it works. $x^2 - 5x + 6 = (x-2)(x-3) = 0 \Rightarrow x = 2$ or $x = 3$.
- Completing the square: always works, reveals the vertex.
- Quadratic formula: always works, use when factorising is hard.
x² − 5x + 6 = 0 を解け。より小さい根はいくつか?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
x² − 5x + 6 = (x − 2)(x − 3) = 0 より、x = 2 または x = 3;小さい方は 2。
Linked equations and modelling
- Quadratics model: projectile paths, profit curves, bridge arches, satellite dishes.
- The vertex gives the maximum (if $a < 0$) or minimum (if $a > 0$) value.
- Example: a profit function $P = -2x^2 + 40x - 150$. Maximum at $x = \dfrac{-40}{2(-2)} = 10$.
- You should solve quadratic equations and quadratic inequalities, and a pair of simultaneous equations (one linear, one quadratic).
You've got it
- complete the square: $a(x+p)^2 + q$, vertex at $(-p, q)$
- the discriminant $b^2 - 4ac$ counts real roots (+ two, 0 one, − none)
- solve with factorising, completing the square, or the quadratic formula
- the vertex gives the maximum or minimum value of the quadratic