가이드대수

이차방정식 및 근호방정식 풀이: 완전한 단계별 가이드

March 26, 2026·18분 읽기·Solvify 팀

이차방정식 및 근호방정식을 푸는 것은 대수에서 가장 중요한 두 가지 기술을 나타내며 대부분의 Algebra 2 교과과정, SAT 수학 및 모든 미적분 전 과정에 함께 나타납니다. 이차방정식은 x²를 최고차항으로 가지며, 근호방정식은 근호 기호 내에 변수가 있습니다. 두 주제는 한 장 이상을 공유합니다. 근호를 제거하기 위해 양변을 제곱하면 거의 항상 다음에 풀어야 할 방정식으로 이차방정식이 생성됩니다. 이 가이드는 이차방정식의 모든 주요 방법(인수분해, 완전제곱, 근의 공식, 그래프)과 근호방정식의 핵심 분리-제곱 기법, 중요한 무관한 해 확인, 그리고 근호방정식이 중간에 이차방정식으로 변하는 매우 일반적인 상황을 다룹니다. 모든 방법은 실수를 사용하여 완전히 풀어진 수치 예제와 함께 표시되어 각 단계를 정확히 따를 수 있습니다.

01이차방정식 및 근호방정식이란 무엇인가?
02이차방정식 풀이: 네 가지 방법
03Solving Radical Equations Step by Step
04When Squaring a Radical Produces a Quadratic
05Common Mistakes and How to Avoid Them
06Practice Problems with Full Solutions
07FAQ — Solving Quadratic and Radical Equations

이차방정식 및 근호방정식이란 무엇인가?

이차방정식은 ax² + bx + c = 0 형태로 쓸 수 있는 방정식입니다(a ≠ 0). 변수의 최고차수는 2입니다. 이차방정식은 양이 비상수 속도로 변하는 곳에 나타나는데, 발사체 운동, 넓이 문제, 피타고라스 정리를 포함하는 기하학 문제에서입니다. y = ax² + bx + c의 그래프는 포물선이며, ax² + bx + c = 0의 실근은 포물선이 x축과 만나는 x값입니다. 교점의 개수는 판별식 D = b² − 4ac에 따라 달라집니다: D > 0이면 두 개의 서로 다른 실근이 있습니다; D = 0이면 정확히 하나의 실근이 있습니다(꼭짓점이 x축에 접합니다); D < 0이면 실근이 없고 해는 복소수입니다. 근호방정식은 근호 부호 내에 변수를 포함합니다—대부분 제곱근(√)이지만 세제곱근과 더 높은 차수의 근호도 있습니다. 예: √(2x + 3) = 5, √(x − 1) = x − 3, ³√(x + 2) = 4. 결정적인 도전은 단순한 대수적 조작만으로는 풀 수 없다는 것입니다—근호를 제거하려면 근호의 지수에 해당하는 거듭제곱으로 양변을 올려야 합니다. 제곱근의 경우 양변을 제곱하는 것을 의미합니다; 세제곱근의 경우 세제곱입니다. 결정적인 복잡성은 양변을 제곱하는 것이 역가능한 연산이 아니라는 것입니다. 3과 −3 모두 9로 제곱되기 때문에 제곱하면 제곱된 방정식을 만족하지만 원래 방정식을 위반하는 해를 도입할 수 있습니다. 이를 무관한 해라고 하며, 근호방정식의 모든 해는 수용되기 전에 원래 방정식에서 검증되어야 합니다. 이 추가 검증 단계는 근호방정식을 대부분의 다른 방정식 유형과 구분하며 평가에서 최대 단일 오류 출처입니다. 두 주제 간의 연결은 직접적입니다: 많은 근호방정식은 제곱 후 그 다음 풀어야 할 이차방정식을 생성합니다. 이차방정식과 근호방정식을 결합된 기술 세트로 푸는 것은 이 전체 문제 클래스를 처음부터 끝까지 다룰 수 있음을 의미합니다.

판별식 규칙: ax² + bx + c = 0에 대해, D = b² − 4ac. D > 0 → 두 개의 실근. D = 0 → 반복된 근. D < 0 → 실근 없음. 모든 근호방정식에 대해: 원래 방정식의 모든 해를 확인하세요—이 단계를 절대 건너뛰지 마세요.

이차방정식 풀이: 네 가지 방법

이차방정식을 푸는 네 가지 표준 방법이 있습니다. 어느 것도 보편적으로 가장 빠르지는 않습니다—각각은 특정 상황에서 최적으로 작동합니다. 먼저 어느 것을 선택해야 하는지 알면 불필요한 계산을 피할 수 있습니다. 네 가지 방법은: (1) 인수분해, 삼항식이 작은 정수 인수를 가질 때 가장 빠름; (2) 완전제곱, 정점 형태가 필요하거나 주계수가 1이고 중간항이 짝수일 때 최적; (3) 근의 공식, 모든 이차방정식에 적용되지만 가장 많은 계산이 필요; 및 (4) 그래프, 근을 추정하거나 대수 해를 확인할 때 유용합니다. 아래의 서로 다른 방정식에서 네 가지 모두 시연되어 각 접근 방식이 가장 잘 작동하는 위치를 보여줍니다.

1. 방법 1: 인수분해 — x² − 7x + 12 = 0 풀기

c와 같은 곱(여기서는 12)과 b와 같은 합(여기서는 −7)을 가지는 두 개의 정수를 찾습니다. 12로 곱해지는 정수 쌍: 1 × 12, 2 × 6, 3 × 4, 그리고 그들의 음수. 이 중에서 −3과 −4는 +12로 곱해지고 −7로 더해집니다. 따라서 x² − 7x + 12 = (x − 3)(x − 4) = 0. 영 제곱의 법칙에 의해 x − 3 = 0 또는 x − 4 = 0이고, x = 3 또는 x = 4입니다. 확인: (3)² − 7(3) + 12 = 9 − 21 + 12 = 0 ✓. (4)² − 7(4) + 12 = 16 − 28 + 12 = 0 ✓. 인수분해는 여기서 가장 빠른 선택입니다—인수 쌍을 보면 약 20초 안에 전체 문제가 완료됩니다. 모든 계수가 작은 정수일 때마다 먼저 인수분해를 시도합니다. 10-15초 안에 정수 인수를 찾을 수 없으면 근의 공식으로 전환하세요.

2. Method 2: Completing the Square — Solve x² + 6x − 7 = 0

Step 1: Move the constant to the right: x² + 6x = 7. Step 2: Find (b/2)² = (6/2)² = 3² = 9. Add 9 to both sides: x² + 6x + 9 = 7 + 9 = 16. Step 3: Factor the left side as a perfect square: (x + 3)² = 16. Step 4: Take the square root of both sides (include ±): x + 3 = ±4. Step 5: Solve for x: x = −3 + 4 = 1 or x = −3 − 4 = −7. Check: (1)² + 6(1) − 7 = 1 + 6 − 7 = 0 ✓. (−7)² + 6(−7) − 7 = 49 − 42 − 7 = 0 ✓. This problem could also be solved quickly by factoring as (x + 7)(x − 1) = 0. Completing the square is shown here to illustrate the procedure. It becomes essential when the discriminant is not a perfect square or when vertex form is the actual goal.

3. Method 3: Quadratic Formula — Solve 2x² − 3x − 2 = 0

The quadratic formula applies to any equation ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) / (2a) Here a = 2, b = −3, c = −2. Step 1: Compute the discriminant: D = (−3)² − 4(2)(−2) = 9 + 16 = 25. Step 2: Since D = 25 > 0, there are two distinct real solutions. Step 3: Apply the formula: x = (−(−3) ± √25) / (2 × 2) = (3 ± 5) / 4. Step 4: Two solutions: x = (3 + 5)/4 = 8/4 = 2 and x = (3 − 5)/4 = −2/4 = −1/2. Check: 2(2)² − 3(2) − 2 = 8 − 6 − 2 = 0 ✓. 2(−1/2)² − 3(−1/2) − 2 = 1/2 + 3/2 − 2 = 0 ✓. The quadratic formula is the go-to choice when factoring is not obvious. It always works, produces exact answers (including irrational ones like (3 + √5)/2), and takes about the same time for any quadratic regardless of how messy the coefficients are.

4. Method 4: Graphing — Solve x² − x − 6 = 0 (conceptual)

Graphing means plotting y = x² − x − 6 and reading the x-intercepts. The parabola crosses the x-axis at x = −2 and x = 3. Verification: (−2)² − (−2) − 6 = 4 + 2 − 6 = 0 ✓. (3)² − 3 − 6 = 9 − 3 − 6 = 0 ✓. Factoring gives the same roots instantly: (x − 3)(x + 2) = 0 → x = 3 or x = −2. Graphing is primarily useful when you need a visual check on algebraic work, when you need to estimate irrational roots to one decimal place, or when a problem asks how many real solutions an equation has (which the discriminant also answers instantly without full solving).

5. Method Selector: When to Use Which

Factoring: try first whenever a = 1 and the constant is a small integer. If no factor pair appears within 15 seconds, move on. Completing the square: use when a = 1 and b is even, or when the problem specifically asks for vertex form or the vertex coordinates of the parabola. Quadratic formula: use when factoring fails, when a ≠ 1 with messy coefficients, or when you need exact irrational roots. Always compute D = b² − 4ac first — if D < 0, there are no real solutions and you can stop immediately. Graphing: use to visualize, estimate, or check — rarely as the primary method on a written algebra exam.

풀기 전에 D = b² − 4ac를 계산합니다. D < 0이면 실근이 없습니다—완료. D ≥ 0이면 방법을 선택합니다: 15초 안에 쌍이 나타나면 인수분해, 그렇지 않으면 공식을 사용합니다. 정점 형태 또는 a = 1이고 b가 짝수이면 완전제곱을 완성합니다.

Solving Radical Equations Step by Step

근호 방정식을 푸는 핵심 절차는 4단계입니다. 한쪽 편에서 근호를 분리하고, 양쪽을 지수에 해당하는 거듭제곱으로 올린 후, 결과 방정식을 풀고, 원래 방정식에서 모든 후보 해를 확인합니다. 확인 단계는 선택 사항이 아닙니다. 무관한 해는 시험 문제에서 흔하며 다른 방법으로는 감지할 수 없습니다. 아래는 증가하는 복잡도의 4가지 예시에서 전체 절차를 보여줍니다. 단순한 제곱근 방정식, 결과 방정식이 선형인 제곱근 방정식, 세제곱근 방정식, 같은 쪽에 두 개의 근호항이 있는 방정식입니다.

1. Step 1 — Always Isolate the Radical First

Move any constants not under the radical to the opposite side before squaring. For √(x − 3) + 5 = 9: subtract 5 first to get √(x − 3) = 4, then square. If you square with the +5 still present, you get (√(x − 3) + 5)² = 81, which expands to x − 3 + 10√(x − 3) + 25 = 81. That is a harder radical equation than the one you started with. Once isolated: √(x − 3) = 4 → square → x − 3 = 16 → x = 19. Check: √(19 − 3) + 5 = √16 + 5 = 4 + 5 = 9 ✓. Always isolate first.

2. Worked Example: Simple Square Root — Solve √(2x + 3) = 5

Step 1: The radical is already isolated. Step 2: Square both sides: (√(2x + 3))² = 5² → 2x + 3 = 25. Step 3: Solve: 2x = 22 → x = 11. Step 4: Check in the original equation: √(2(11) + 3) = √(22 + 3) = √25 = 5 ✓. Final answer: x = 11. One solution, no extraneous issue. This is the simplest case: after squaring, you get a linear equation with exactly one solution.

3. Worked Example: Cube Root — Solve ³√(x − 5) = 3

For a cube root, cube both sides (raise to the 3rd power) rather than squaring. Step 1: The radical is already isolated. Step 2: Cube both sides: (³√(x − 5))³ = 3³ → x − 5 = 27. Step 3: Solve: x = 32. Step 4: Check: ³√(32 − 5) = ³√27 = 3 ✓. Cube root equations rarely produce extraneous solutions because cubing is a one-to-one operation — no two distinct real numbers cube to the same value. Even so, checking is still good practice. General rule: for a radical with index n, raise both sides to the nth power. √ → square (power 2), ³√ → cube (power 3), ⁴√ → raise to the 4th power.

4. Worked Example: Two Radicals Equal — Solve √(3x + 1) = √(x + 9)

When both sides are square roots set equal to each other, squaring both sides eliminates both radicals at once. Step 1: The equation is ready to square. Step 2: Square: 3x + 1 = x + 9. Step 3: Solve: 2x = 8 → x = 4. Step 4: Check in the original: left = √(3(4) + 1) = √13. Right = √(4 + 9) = √13 ✓. Final answer: x = 4. Even when two-radical equations produce only one candidate, always check it — not all single-candidate equations are guaranteed to be valid.

The four steps for every radical equation: (1) isolate the radical, (2) raise both sides to the power matching the index, (3) solve the resulting equation, (4) check every solution in the original. Step 4 is mandatory — extraneous solutions cannot be detected any other way.

When Squaring a Radical Produces a Quadratic

Algebra 2 시험에서 가장 자주 시험되는 시나리오는 우변이 x의 일차 또는 이차 식인 근호 방정식입니다. 제곱한 후, 이를 풀어야 하는 이차 방정식을 얻게 되며, 두 근 모두 무관한 해에 대해 확인해야 합니다. 이것은 이차 방정식과 근호 방정식을 푸는 것이 직접 겹치는 곳입니다. 아래의 3가지 완전히 풀어진 예는 3가지 주요 형식을 다룹니다: 근호가 일차 단항식과 같은 경우(√ = x), 근호가 이항식과 같은 경우(√ = x + n), 및 피개방수 자체에 x²이 포함된 경우입니다.

1. Example 1 — √(x + 6) = x (radical equals a linear term)

Step 1: The radical is isolated. Step 2: Square both sides: x + 6 = x². Step 3: Rearrange into standard form: x² − x − 6 = 0. Step 4: Factor: (x − 3)(x + 2) = 0. Candidates: x = 3 or x = −2. Step 5: Check in the original √(x + 6) = x: x = 3: √(3 + 6) = √9 = 3. Right side = 3 ✓. Valid. x = −2: √(−2 + 6) = √4 = 2. Right side = −2. Since 2 ≠ −2, this is extraneous — reject. Final answer: x = 3 only. The value x = −2 is extraneous because √ always denotes the principal (non-negative) square root, which can never equal a negative number.

2. Example 2 — √(2x + 9) = x + 3 (radical equals a binomial)

Step 1: The radical is isolated. Step 2: Square both sides: 2x + 9 = (x + 3)² = x² + 6x + 9. Step 3: Rearrange: x² + 6x + 9 − 2x − 9 = 0 → x² + 4x = 0. Step 4: Factor: x(x + 4) = 0. Candidates: x = 0 or x = −4. Step 5: Check in the original √(2x + 9) = x + 3: x = 0: √(0 + 9) = √9 = 3. Right side = 0 + 3 = 3 ✓. Valid. x = −4: √(2(−4) + 9) = √(−8 + 9) = √1 = 1. Right side = −4 + 3 = −1. Since 1 ≠ −1, extraneous — reject. Final answer: x = 0 only. Again, the extraneous root appears because the right side becomes negative at x = −4, which is impossible for a square root. This pattern — one valid root, one extraneous — is the most common outcome when the right side is a binomial.

3. Example 3 — √(x² − 4) = x − 1 (radicand already quadratic)

Step 1: The radical is isolated. Step 2: Square both sides: x² − 4 = (x − 1)² = x² − 2x + 1. Step 3: The x² terms cancel: −4 = −2x + 1 → −5 = −2x → x = 5/2. Step 4: Only one candidate: x = 5/2. Step 5: Check in the original √(x² − 4) = x − 1: x = 5/2: left = √((5/2)² − 4) = √(25/4 − 16/4) = √(9/4) = 3/2. Right = 5/2 − 1 = 3/2 ✓. Final answer: x = 5/2. Even though the radicand was already quadratic, the x² terms cancelled after squaring, leaving a linear equation with one solution. This is not always predictable — always work through the algebra fully rather than assuming the degree of the result.

When the right side of a radical equation is a binomial (like x − 2 or x + 3), squaring gives (x ± n)² on the right — expand it fully. The resulting quadratic will almost always have two roots, but typically only one survives the extraneous-solution check. Never assume both are valid.

Common Mistakes and How to Avoid Them

Specific, repeated errors account for most lost marks on quadratic and radical equation problems. The five mistakes below cover both equation types. Each is paired with a concrete correction so you can calibrate your technique before the next assessment.

1. Mistake 1 — Skipping the extraneous solution check (radical equations)

This is the most frequent and costly error. After solving √(x + 4) = x − 2, students obtain two algebraic roots (x = 0 and x = 5) and stop there. But at x = 0, the right side is 0 − 2 = −2 < 0, which is impossible for a square root. Only x = 5 is valid. Fix: after solving the squared equation, substitute every candidate back into the original equation (with the radical sign) and reject any that make the equation false. There is no algebraic shortcut for this — you must substitute.

2. Mistake 2 — Squaring before isolating the radical

For √(x − 3) + 5 = 9, squaring both sides immediately gives (√(x − 3) + 5)² = 81, which expands to x − 3 + 10√(x − 3) + 25 = 81 — a new, harder radical equation. Fix: subtract 5 from both sides first to get √(x − 3) = 4. Then square: x − 3 = 16 → x = 19. Check: √(19 − 3) + 5 = 4 + 5 = 9 ✓. Isolating the radical first always makes the squaring step cleaner.

3. Mistake 3 — Expanding a binomial square incorrectly

A very common algebra error when squaring the right side: writing (x − 2)² = x² − 4 instead of x² − 4x + 4. The middle term 2ab is forgotten or miscalculated, which changes the quadratic you obtain and leads to wrong roots. Fix: always use (a − b)² = a² − 2ab + b². For (x − 2)²: a = x, b = 2, so (x)² − 2(x)(2) + (2)² = x² − 4x + 4. Write out all three terms. The middle term is 2 × x × 2 = 4x — write it explicitly before simplifying.

4. Mistake 4 — Sign error in the discriminant (quadratic formula)

For 2x² − 3x − 2 = 0 with a = 2, b = −3, c = −2: the discriminant is D = (−3)² − 4(2)(−2). Students frequently compute 9 − 8 = 1 instead of 9 + 16 = 25 because they drop the negative on c, treating −4(2)(−2) as if it were −4(2)(2). Fix: write out the substitution with explicit parentheses: D = (−3)² − 4(2)(−2) = 9 − (−16) = 9 + 16 = 25. When c is negative, the term −4ac becomes positive. Parentheses around the substituted value prevent sign errors.

5. Mistake 5 — Omitting ± when taking a square root

After completing the square and arriving at (x + 3)² = 16, many students write x + 3 = 4 and find only x = 1, missing x = −7. Fix: every time you take a square root to solve an equation (not read a radical sign in the original), write ±. The equation (x + 3)² = 16 gives x + 3 = ±4 → x = 1 or x = −7. The ± is where both solutions come from — omitting it always discards one root. This is distinct from radical equations: when the original equation has √ on the left, the radical denotes only the positive root, and the second solution appears only through the extraneous check.

Two rules that prevent the majority of errors: (1) every radical equation solution requires a substitution check in the original. (2) every square root taken during solving produces ±, not just +. Both rules protect you from losing valid solutions or accepting invalid ones.

Practice Problems with Full Solutions

Five problems cover the full range of skills involved in solving quadratic and radical equations. Problems 1 and 2 are pure quadratic equations using factoring and the formula. Problems 3 and 4 are radical equations — one clean, one with an extraneous solution. Problem 5 is a mixed radical-quadratic equation where both roots survive the check. Work each problem fully before reading the solution.

1. Problem 1 (Quadratic — Factoring) — Solve x² + 2x − 15 = 0

Look for two integers with product −15 and sum +2. Options: (1, −15), (−1, 15), (3, −5), (−3, 5). The pair (5, −3) multiplies to −15 and adds to 2. So x² + 2x − 15 = (x + 5)(x − 3) = 0. Solutions: x = −5 or x = 3. Check: (−5)² + 2(−5) − 15 = 25 − 10 − 15 = 0 ✓. (3)² + 2(3) − 15 = 9 + 6 − 15 = 0 ✓.

2. Problem 2 (Quadratic — Formula) — Solve 3x² + 5x − 1 = 0

Here a = 3, b = 5, c = −1. Factoring is not practical because there is no clean integer factor pair. Step 1: D = b² − 4ac = (5)² − 4(3)(−1) = 25 + 12 = 37. Step 2: D = 37 > 0, so two distinct real solutions exist. √37 is not a perfect square, so the roots are irrational. Step 3: x = (−5 ± √37) / 6. Solutions: x = (−5 + √37) / 6 ≈ (−5 + 6.083) / 6 ≈ 0.181 and x = (−5 − √37) / 6 ≈ −1.847. Vieta's check: sum of roots = −b/a = −5/3 ≈ −1.667. Computed sum: 0.181 + (−1.847) ≈ −1.666 ✓. Product of roots = c/a = −1/3 ≈ −0.333. Computed product: 0.181 × (−1.847) ≈ −0.334 ✓.

3. Problem 3 (Radical — No Extraneous Solution) — Solve √(3x − 2) = 4

Step 1: The radical is already isolated. Step 2: Square both sides: 3x − 2 = 16. Step 3: Solve: 3x = 18 → x = 6. Step 4: Check in the original: √(3(6) − 2) = √(18 − 2) = √16 = 4 ✓. Final answer: x = 6.

4. Problem 4 (Radical — Extraneous Solution Present) — Solve √(x + 12) = x

Step 1: The radical is isolated. Step 2: Square both sides: x + 12 = x². Step 3: Rearrange: x² − x − 12 = 0. Step 4: Factor: (x − 4)(x + 3) = 0. Candidates: x = 4 or x = −3. Step 5: Check in the original √(x + 12) = x: x = 4: √(4 + 12) = √16 = 4. Right side = 4 ✓. Valid. x = −3: √(−3 + 12) = √9 = 3. Right side = −3. Since 3 ≠ −3, extraneous — reject. Final answer: x = 4 only. This is a classic example: two algebraic roots, one real, one extraneous.

5. Problem 5 (Radical–Quadratic, Both Roots Valid) — Solve √(x² + 3x) = 2

Step 1: The radical is isolated. Step 2: Square both sides: x² + 3x = 4. Step 3: Rearrange: x² + 3x − 4 = 0. Step 4: Factor: (x + 4)(x − 1) = 0. Candidates: x = −4 or x = 1. Step 5: Check in the original √(x² + 3x) = 2: x = −4: √((−4)² + 3(−4)) = √(16 − 12) = √4 = 2 ✓. Valid. x = 1: √(1² + 3(1)) = √(1 + 3) = √4 = 2 ✓. Valid. Final answer: x = −4 or x = 1. Both solutions are valid — this is less common but fully possible. Both values of x give a radicand of 4, and neither makes a √ equal to a negative number, so neither is extraneous.

FAQ — Solving Quadratic and Radical Equations

These are the questions that come up most frequently when students work through this material. Each answer focuses on the specific mechanical or conceptual point most likely to cause errors.

1. What is an extraneous solution and why does it appear?

An extraneous solution is a value that satisfies the equation after squaring but not the original radical equation. It appears because squaring is not reversible: if the original equation had √(expression) = −5, that is already impossible since square roots are ≥ 0 — but squaring eliminates that impossibility, giving expression = 25, which can have a solution. The squaring step erased the sign constraint. The only way to detect extraneous solutions is to substitute each candidate into the original equation (the one with the radical) and reject any that fail. There is no algebraic shortcut. On exams, problems with radical equations are often designed specifically so that one root is extraneous — always check.

2. Which method should I use to solve a quadratic equation?

Scan for factoring first: look for two integers (or rationals) that multiply to ac and add to b. If you cannot find them in 15 seconds, compute D = b² − 4ac. If D is a perfect square, factoring works and you can try again; if not, the roots are irrational and the quadratic formula is the right tool. If the problem asks for vertex form or the vertex of the parabola, use completing the square. If it asks for the number of real solutions, you need only compute D — no full solving required.

3. Can a radical equation have no solution at all?

Yes — in two distinct ways. First, the equation can be immediately impossible: √(x + 1) = −3 has no solution because a square root is always ≥ 0 and can never equal −3. Second, all algebraic candidates can turn out to be extraneous after checking. Example: solve √(x + 2) = x − 4. Squaring: x + 2 = x² − 8x + 16 → x² − 9x + 14 = 0 → (x − 2)(x − 7) = 0. Check x = 2: √4 = 2 but right side = 2 − 4 = −2. Extraneous. Check x = 7: √9 = 3 and right side = 7 − 4 = 3 ✓. Valid. Here one solution survives, but if both had been extraneous, the equation would have no real solution.

4. Does the quadratic formula work for every quadratic?

Yes, without exception. The formula x = (−b ± √(b² − 4ac)) / (2a) gives the correct solutions for any ax² + bx + c = 0 as long as a ≠ 0. When D < 0, the solutions are complex: x = (−b ± i√(4ac − b²)) / (2a). In a standard Algebra 2 course, you typically note 'no real solutions' and stop. When D = 0, the formula still works — it gives x = −b/(2a) twice, confirming the single repeated root. The formula always applies; use it as the reliable fallback whenever factoring fails.

5. How do I solve a radical equation that has two separate radicals?

When an equation contains two radical terms, isolate one radical and square. If the second radical remains, isolate it and square again. Example: solve √(x + 5) − √(x − 3) = 2. Step 1: Isolate one radical: √(x + 5) = √(x − 3) + 2. Step 2: Square: x + 5 = (x − 3) + 4√(x − 3) + 4 = x + 1 + 4√(x − 3). Step 3: Simplify: 5 − 1 = 4√(x − 3) → 4 = 4√(x − 3) → √(x − 3) = 1. Step 4: Square again: x − 3 = 1 → x = 4. Step 5: Check in the original: √(4 + 5) − √(4 − 3) = √9 − √1 = 3 − 1 = 2 ✓. Final answer: x = 4. Two-radical equations almost always require two rounds of squaring and always require a final check.

6. How do I know how many real solutions a quadratic has without fully solving it?

Compute the discriminant D = b² − 4ac and read the result directly: D > 0 → two distinct real solutions (parabola crosses x-axis twice). D = 0 → one repeated real solution (vertex touches x-axis). D < 0 → no real solutions (parabola does not cross x-axis). Example: how many real solutions does 2x² − 4x + 3 = 0 have? D = (−4)² − 4(2)(3) = 16 − 24 = −8 < 0. Answer: no real solutions — without any further work. This is the fastest approach to 'how many solutions' questions on multiple-choice tests.

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