指南代數

如何使用二次方程 — 逐步指南

March 14, 2026·14分鐘閱讀·Solvify Team

二次方程是代數中最實用的工具之一，一旦您知道如何應用它，沒有任何二次方程能夠難住您。每個二次方程都符合標準形式 ax² + bx + c = 0，二次公式 x = (−b ± √(b² − 4ac)) / 2a 在一次計算中給出兩個解。如果您曾經在搜索欄中輸入過「如何使用二次方程」，本指南就是答案 — 它涵蓋從識別係數到檢查最終答案的每一步，全程配有實際的解答示例。

什麼是二次方程？

二次方程是任何多項式方程，其中變量的最高次冪是2。標準形式是 ax² + bx + c = 0，其中 a、b 和 c 是實數，a 不能等於零 — 如果 a 為零，x² 項將消失，方程將變成線性的。「二次」一詞來自拉丁文 quadratus，意思是「平方」，因為定義特徵總是平方變量。二次方程到處都是：拋出球的弧線遵循二次路徑，企業的利潤曲線通常是二次的，電路的共振頻率通過求解二次方程找到。因此，了解如何使用二次公式是一項超越課堂的真正技能。有三種常見的求解二次方程方法 — 因式分解、配方法和二次公式。當因式分解有效時很快，但許多二次方程在整數上不能整齊分解。二次公式總是有效的，對於每個具有實根或複根的二次方程都適用，這就是為什麼值得冷硬記住它。在深入機制之前，請注意，每個「逐步教我如何使用二次方程」的請求通常歸結為一個潛在問題：我如何可靠地從混亂的方程到正確的數值答案？答案是一個可重複的六步程序。

Standard form: ax² + bx + c = 0, where a ≠ 0. The quadratic formula: x = (−b ± √(b² − 4ac)) / 2a.

識別 a、b 和 c — 每次的第一步

在您能將任何值代入二次公式之前，您需要正確讀取方程式並提取三個係數。係數 a 屬於 x² 項，b 屬於 x 項，c 是沒有變數的常數。如果某一項缺失，其係數為零 — 例如，x² − 9 = 0 沒有 x 項，所以 b = 0。正確獲得這些值是後續所有操作的基礎，誤讀符號是目前最常見的錯誤來源。在識別 a、b 和 c 之前，始終將方程式改寫為標準形式 — 左側所有項，右側為零。您在此步驟上花費的三十秒可以防止最昂貴的代數錯誤。

1. Move all terms to one side so the equation equals zero

Example: 3x² = 7x − 2 must become 3x² − 7x + 2 = 0 before you do anything else. Subtract 7x and add 2 to both sides. The equation must equal zero for the quadratic formula to apply.

2. Read off a — the coefficient of x²

In 3x² − 7x + 2 = 0, a = 3. If the equation reads x² − 5x + 4 = 0, there is an invisible 1 in front, so a = 1. Never skip writing a = 1 explicitly; it prevents errors later when you compute 2a.

3. Read off b — the coefficient of x (sign included)

In 3x² − 7x + 2 = 0, b = −7, not +7. The minus sign is part of b. Students who write b = 7 and then try to remember the sign later consistently make errors. Write the full signed value.

4. Read off c — the constant term

In 3x² − 7x + 2 = 0, c = 2. If there is no constant term (e.g., 3x² − 7x = 0), then c = 0. Again, write it down explicitly rather than carrying it in your head.

5. Write a, b, c beside the equation before proceeding

Label them: a = 3, b = −7, c = 2. This takes ten seconds and gives you a reference point for every subsequent calculation. It also makes it easy to find your mistake if the check step fails.

如何使用二次方程 — 完整分步指南

以下是完整的方法 — 「如何使用二次方程」的完整答案。二次公式是 x = (−b ± √(b² − 4ac)) / 2a。± 符號表示您計算兩個答案：一個使用加法（+ 的情況）和一個使用減法（− 的情況）。兩個答案都是方程的有效解。首先進行一個清晰的示例：x² + 5x + 6 = 0。識別：a = 1，b = 5，c = 6。按順序執行每個步驟，不要跳過。

1. Step 1 — Write the quadratic formula

Always begin by writing x = (−b ± √(b² − 4ac)) / 2a on your paper before substituting anything. This gives you a template and makes the structure visible. It also prevents the common mistake of forgetting part of the formula under exam pressure.

2. Step 2 — Compute −b

b = 5, so −b = −5. In this example it is simple, but forming the habit of treating −b as a separate computation pays off when b is negative — e.g., if b = −3, then −b = +3.

3. Step 3 — Compute the discriminant b² − 4ac

b² = 5² = 25. Then 4ac = 4 × 1 × 6 = 24. The discriminant is b² − 4ac = 25 − 24 = 1. A positive discriminant means two distinct real solutions. Write this value down before moving on.

4. Step 4 — Take the square root of the discriminant

√1 = 1. This is a perfect square, so you get clean integer answers. If the discriminant had been, say, 12, you would simplify √12 = 2√3 before proceeding.

5. Step 5 — Compute 2a

2a = 2 × 1 = 2. This is the denominator for both solutions. Write it separately so you do not accidentally divide only part of the numerator.

6. Step 6 — Find both solutions using + and −

x = (−5 + 1) / 2 = −4 / 2 = −2. And x = (−5 − 1) / 2 = −6 / 2 = −3. The two solutions are x = −2 and x = −3. Write both.

7. Step 7 — Check your answers by substituting back

Check x = −2: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0 ✓. Check x = −3: (−3)² + 5(−3) + 6 = 9 − 15 + 6 = 0 ✓. Both solutions satisfy the original equation. The check step is not optional — it is the only reliable way to catch arithmetic errors.

The quadratic formula x = (−b ± √(b² − 4ac)) / 2a works for every quadratic equation. The ± always produces two solutions — write both.

完成前理解判別式

根號下的表達式 — b² − 4ac — 稱為判別式。值得在完成公式的其餘部分之前先計算這個單一值，因為它立即告訴您可以預期什麼樣的解。如果判別式為負，對於標準代數課程，您可以在這裡停止（沒有實數解）。如果它為零，您已經知道有一個重根。如果它是完全平方數，您可以期望得到簡潔的有理答案。首先檢查判別式是一個五秒的小投資，可以為您節省一分鐘的徒勞無功的運算。

1. Discriminant > 0 — two distinct real solutions

The equation crosses the x-axis at two points. Example: x² − 5x + 4 = 0 has discriminant 25 − 16 = 9. √9 = 3. Solutions: x = (5 + 3)/2 = 4 and x = (5 − 3)/2 = 1.

2. Discriminant = 0 — exactly one real solution (repeated root)

The parabola just touches the x-axis at its vertex. Example: x² − 6x + 9 = 0 has discriminant 36 − 36 = 0. Solution: x = 6/2 = 3 only. This is called a double root — the same answer appears twice.

3. Discriminant < 0 — no real solutions

The parabola does not cross the x-axis. Example: x² + 2x + 5 = 0 has discriminant 4 − 20 = −16. There are no real solutions. In complex-number algebra the solutions are x = −1 ± 2i, but in a standard high school course the answer is 'no real solution.'

b² − 4ac > 0 → two real roots. b² − 4ac = 0 → one repeated root. b² − 4ac < 0 → no real roots.

如何使用二次方程 — 一個更難的例子

現在讓我們將同樣的過程應用於具有負 b 的問題 — 導致最多符號錯誤的類型。問題：2x² − 3x − 5 = 0。識別：a = 2，b = −3，c = −5。在每個對符號敏感的步驟中注意。

1. Write a, b, c explicitly

a = 2, b = −3, c = −5. Note that both b and c are negative. Write these values labeled before touching the formula.

2. Compute −b

b = −3, so −b = −(−3) = +3. This is a critical step: flipping the sign of a negative b gives a positive result. Students who skip this sub-step and write −(−3) incorrectly in the heat of an exam lose easy marks.

3. Compute the discriminant b² − 4ac

b² = (−3)² = 9. Note: squaring a negative number gives a positive result — (−3)² = 9, not −9. Then 4ac = 4 × 2 × (−5) = −40. So b² − 4ac = 9 − (−40) = 9 + 40 = 49. Subtracting a negative is the same as adding.

4. Take the square root of the discriminant

√49 = 7. This is a perfect square, so the answers will be rational. Good sign — factoring might have worked here too.

5. Compute 2a

2a = 2 × 2 = 4.

6. Find both solutions

x = (3 + 7) / 4 = 10 / 4 = 5/2 = 2.5. And x = (3 − 7) / 4 = −4 / 4 = −1. The solutions are x = 2.5 and x = −1.

7. Check both solutions

For x = 2.5: 2(2.5)² − 3(2.5) − 5 = 2(6.25) − 7.5 − 5 = 12.5 − 7.5 − 5 = 0 ✓. For x = −1: 2(−1)² − 3(−1) − 5 = 2 + 3 − 5 = 0 ✓. Both check out.

When b is negative, −b becomes positive. When c is negative, subtracting 4ac adds to the discriminant. Track every sign change as its own computation.

完整解答的練習題

在閱讀解決方案之前，自己先進行每個問題。首先識別 a、b 和 c，並寫出判別式。以下五個問題涵蓋您在測試中會遇到的所有情況。

1. Problem 1 — Easy: x² − 7x + 12 = 0

a = 1, b = −7, c = 12. Discriminant: (−7)² − 4(1)(12) = 49 − 48 = 1. √1 = 1. x = (7 + 1)/2 = 8/2 = 4 and x = (7 − 1)/2 = 6/2 = 3. Solutions: x = 4 and x = 3. Check: 16 − 28 + 12 = 0 ✓ and 9 − 21 + 12 = 0 ✓.

2. Problem 2 — Medium: 3x² + 10x + 3 = 0

a = 3, b = 10, c = 3. Discriminant: 100 − 36 = 64. √64 = 8. x = (−10 + 8)/6 = −2/6 = −1/3 and x = (−10 − 8)/6 = −18/6 = −3. Solutions: x = −1/3 and x = −3. Check for x = −3: 3(9) + 10(−3) + 3 = 27 − 30 + 3 = 0 ✓.

3. Problem 3 — Repeated root: 4x² − 4x + 1 = 0

a = 4, b = −4, c = 1. Discriminant: 16 − 16 = 0. One repeated root: x = 4 / 8 = 1/2. Solution: x = 1/2 only. Check: 4(1/4) − 4(1/2) + 1 = 1 − 2 + 1 = 0 ✓.

4. Problem 4 — Hard: 5x² + 2x − 7 = 0

a = 5, b = 2, c = −7. Discriminant: 4 − 4(5)(−7) = 4 + 140 = 144. √144 = 12. x = (−2 + 12)/10 = 10/10 = 1 and x = (−2 − 12)/10 = −14/10 = −7/5. Solutions: x = 1 and x = −1.4. Check for x = 1: 5 + 2 − 7 = 0 ✓.

5. Problem 5 — Applied: A ball is thrown upward with height h = −16t² + 64t + 80 feet. When does it hit the ground?

Set h = 0: −16t² + 64t + 80 = 0. Divide through by −16: t² − 4t − 5 = 0. a = 1, b = −4, c = −5. Discriminant: 16 + 20 = 36. √36 = 6. t = (4 + 6)/2 = 5 and t = (4 − 6)/2 = −1. Since time cannot be negative, discard t = −1. The ball hits the ground at t = 5 seconds.

常見錯誤及修復方法

這七個錯誤造成二次方程問題上大多數失分。即使您感到自信，也請閱讀它們 — 每個都有一個具體的、可操作的修復方法，您可以在下次測試前應用。

1. Not converting to standard form first

The quadratic formula requires the equation to equal zero. For 2x² + 3 = 5x, students sometimes read a = 2, b = 3, c = 5 and get a completely wrong answer. Always rewrite as 2x² − 5x + 3 = 0 first. Then a = 2, b = −5, c = 3.

2. Misreading the sign of b

If the equation has −5x, then b = −5. The minus sign is not separate from b — it belongs to it. Writing b = 5 and then 'remembering' the negative later guarantees errors. Write the full signed value: b = −5.

3. Squaring a negative b incorrectly

(−5)² = 25, not −25. Squaring always produces a non-negative result. This is the most common single-step error with the quadratic formula. Use parentheses: always write (b)² and substitute the signed value inside them.

4. Writing only one solution instead of two

The ± means you must write two answers. If you only write the + case, you are missing a solution. Even on a multiple-choice test, both solutions matter — the problem may be looking for the larger root, the smaller root, or both.

5. Dividing only part of the numerator by 2a

The formula is (−b ± √(b²−4ac)) / 2a. Both −b and the ±√ part must be divided by 2a. A frequent error is writing −b ± √(b²−4ac)/2a, which divides only the radical. Draw a long fraction bar under the entire numerator.

6. Arithmetic errors inside the square root

√(b² − 4ac) cannot be split into √b² − √(4ac). You must compute the full numerical value under the radical first (b² − 4ac = some number), and then take the square root of that one number. Compute it as a separate sub-problem.

7. Skipping the check step

Substituting both answers back into the original equation takes thirty seconds and catches every sign and arithmetic mistake. If a solution does not check out, go back to the discriminant step and find the error. Do not turn in unchecked answers.

何時使用二次公式與其他方法

二次公式總是有效的 — 它是通用的備選方案。但在某些情況下，其他方法更快。當方程有小整數根時，因數分解花費不到一分鐘。完成平方法在推導拋物線的頂點形式時很有用。使用判別式指導您的選擇：如果 b² − 4ac 是完全平方數（1、4、9、16、25、36、49、64、81、100、121、144...），根是有理數，因數分解可能更快。如果它不是完全平方數，直接跳到二次公式 — 您需要十進制或根式答案，無論如何，有理數上的因數分解都行不通。在測試壓力下，許多學生在前幾個問題之後對所有問題都預設使用二次公式。這是一個完全合理的策略：它花費的時間比因數分解稍長，但它從不失敗，一旦您自動化該方法，它很少會產生符號錯誤。

Quick decision rule: if b² − 4ac is a perfect square, try factoring. Otherwise, use the quadratic formula directly.

獲得更快、更可靠結果的提示

一旦核心方法成為自動化，這些習慣就將始終獲得滿分的學生與每道題丟失一兩分的學生區分開來。

1. Memorize the formula correctly — write it from scratch each time

Do not look up the quadratic formula mid-problem. Memorize x = (−b ± √(b² − 4ac)) / 2a and write it from memory at the top of your work before you substitute. The act of writing it focuses your attention and provides a reference template.

2. Compute the discriminant as a dedicated sub-problem

Calculate b² − 4ac and box the answer before proceeding. Label it as the discriminant. This one habit eliminates about half of all quadratic formula errors, because students who compute b² and 4ac separately are much less likely to mix up signs.

3. Put parentheses around every substituted value

Write (−3)² not −3². Write 4(2)(−5) not 4 × 2 × −5. Parentheses force correct order of operations and catch sign errors before they propagate.

4. Simplify the square root before dividing by 2a

If the discriminant is 48, write √48 = √(16 × 3) = 4√3 before dividing by 2a. Simplifying first produces smaller numbers to work with and gives neater final answers.

5. Use Vieta's formulas as a fast sanity check

The sum of the two roots equals −b/a, and their product equals c/a. For any quadratic ax² + bx + c = 0, verify these relations before writing your final answer. Example: for x² + 5x + 6 = 0 with roots −2 and −3: sum = −2 + (−3) = −5 = −5/1 ✓, product = (−2)(−3) = 6 = 6/1 ✓. If these fail, recheck your arithmetic.

6. For decimal answers, keep at least two decimal places

Unless the problem asks for exact radical form, round to two decimal places and double-check by substitution. For 5x² + 2x − 7 = 0, x = 1 checks cleanly; x = −1.40 gives 5(1.96) + 2(−1.40) − 7 = 9.8 − 2.8 − 7 = 0 ✓.

常見問題 — 如何使用二次方程

這些是學生首次學習應用二次公式時最常提出的問題。其中許多是「在這個特定情況下如何使用二次方程」的變體。每個答案都側重於實際操作而不是理論。

1. What if a is a negative number?

The formula still works exactly the same way. Substitute the negative value for a. For example, if a = −2, then 2a = −4, and your solutions are divided by −4. Be particularly careful with the discriminant: 4ac with a negative a means you are computing 4 × (negative) × c, which reverses the sign of that term.

2. Can the quadratic formula always be used, or only sometimes?

It can always be used for any quadratic equation ax² + bx + c = 0 where a ≠ 0. Unlike factoring, which fails when roots are irrational, the quadratic formula handles every case — integer roots, fractional roots, irrational roots (involving √), and complex roots. If you can only memorize one method, make it the quadratic formula.

3. What does it mean when I get a negative number under the square root?

When b² − 4ac < 0, there are no real solutions. In a standard pre-calculus or algebra 2 course, the expected answer is 'no real solutions.' In a complex numbers unit, you write the solutions using i = √(−1): x = (−b ± i√(4ac − b²)) / 2a. Which answer is expected depends on your course level.

4. My two solutions have opposite signs. Is that normal?

Yes, completely normal. When c is negative (e.g., ax² + bx − 5 = 0), the product of the two roots equals c/a, which is negative. For the product of two numbers to be negative, one must be positive and the other negative. So when c < 0, you can expect one positive and one negative solution.

5. How do I handle a quadratic with no x term (b = 0)?

If b = 0, the equation is ax² + c = 0. The quadratic formula simplifies to x = ±√(−c/a). For example, 2x² − 8 = 0 gives x = ±√(8/2) = ±√4 = ±2. You could also solve this by isolating x²: x² = 4, so x = ±2. Both approaches give the same result.

6. What is the relationship between the quadratic formula and completing the square?

The quadratic formula is derived by completing the square on the general equation ax² + bx + c = 0. They are the same method — the formula is just what completing the square looks like when applied to a general a, b, c rather than specific numbers. If you understand completing the square, you can re-derive the formula any time you forget it.

7. Should I leave answers as exact fractions or convert to decimals?

Check what the problem asks for. Applied problems (rates, distances, times) usually want decimals rounded to a stated precision. Pure algebra problems typically want exact answers: fractions, radicals, or integers. When in doubt, give the exact answer and a decimal approximation side by side, e.g., x = (3 + √5)/2 ≈ 2.618.

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