Lösa Kvadratiska och Radikala Ekvationer: Komplett Steg-för-Steg Guide
Att lösa kvadratiska och radikala ekvationer representerar två av de viktigaste färdigheterna inom algebra — och de förekommer tillsammans i de flesta Algebra 2-kursplaner, på SAT Math och i varje precalculus-kurs. En kvadratisk ekvation har x² som sin högsta gradterm; en radikal ekvation har variabeln innanför ett rottecken. De två ämnena delar mer än ett kapitel: att kvadrera båda sidorna för att eliminera en radikal producerar nästan alltid en kvadratisk ekvation som nästa att lösa. Den här guiden täcker varje huvudmetod för kvadratiska — faktorisering, att fylla i kvadraten, kvadratformeln och grafik — plus den centrala isolerings-och-kvadrerings-tekniken för radikala ekvationer, den kritiska kontrollen av främmande lösningar och den mycket vanliga situationen där en radikal ekvation blir kvadratisk på vägen. Varje metod visas med ett helt utarbetat numeriskt exempel med reella tal så att du kan följa varje steg exakt.
Innehåll
Vad är kvadratiska och radikala ekvationer?
En kvadratisk ekvation är vilken ekvation som helst som kan skrivas i standardform som ax² + bx + c = 0, där a ≠ 0. Den högsta potensen av variabeln är 2. Kvadratiska ekvationer visas överallt där en kvantitet ändras i en konstant hastighet — i projektilrörelse, områdesproblem och geometrifrågor som involverar Pythagoras sats. Grafen för y = ax² + bx + c är en parabel, och de verkliga lösningarna av ax² + bx + c = 0 är x-värdena där parabeln korsar x-axeln. Hur många korsningar som finns beror på diskriminanten D = b² − 4ac: om D > 0, finns det två olika reella rötter; om D = 0, finns det exakt en reell rot (vertexen berör x-axeln); om D < 0, finns det inga reella rötter och lösningarna är komplexa tal. En radikal ekvation innehåller en variabel inuti ett radikalt tecken — oftast en kvadratrot (√), även om kubroter och högre radikal också existerar. Exempel: √(2x + 3) = 5, √(x − 1) = x − 3, ³√(x + 2) = 4. Den definierande utmaningen är att du inte kan lösa dessa genom enkel algebraisk manipulation — du måste höja båda sidor till potensen som motsvarar det radikala indexet för att eliminera roten. För en kvadratrot betyder det att kvadrera båda sidor; för en kubrot, kuba. Den avgörande komplikationen är att kvadrera båda sidor inte är en reversibel operation. Eftersom både 3 och −3 kvadreras till 9, kan kvadrering introducera lösningar som uppfyller den kvadrerade ekvationen men bryter mot originalsekvationen. Dessa kallas främmande lösningar, och varje lösning av en radikal ekvation måste verifieras i den ursprungliga ekvationen innan den accepteras. Detta extra verifieringssteg skiljer radikala ekvationer från de flesta andra ekvationstyper och är den enskilt största felkällan vid bedömningen. Kopplingen mellan de två ämnena är direkt: många radikala ekvationer producerar, efter kvadrering, en kvadratisk som då måste lösas. Att lösa kvadratiska och radikala ekvationer som en kombinerad färdighetsuppsättning betyder att du kan hantera denna helt klass av problem från början till slut.
Diskriminantregel: för ax² + bx + c = 0, D = b² − 4ac. D > 0 → två reella rötter. D = 0 → en upprepad rot. D < 0 → inga reella rötter. För varje radikal ekvation: verifiera ALLA lösningar i originalet — hoppa aldrig över detta steg.
Lösa kvadratiska ekvationer: fyra metoder
Det finns fyra standardmetoder för att lösa en kvadratisk ekvation. Ingen är universellt snabbast — var och en fungerar bäst i specifika situationer. Att veta vilken du ska välja först undviker onödig aritmetik. De fyra metoderna är: (1) faktorisering, snabbast när trinomiet har små heltalsfaktorer; (2) slutföra kvadraten, bäst när vertexform behövs eller den ledande koefficienten är 1 med en jämn mittterm; (3) den kvadratiska formeln, som fungerar för varje kvadratisk men involverar mest beräkning; och (4) grafik, användbar för att uppskatta rötter eller verifiera algebraiska lösningar. Alla fyra demonstreras nedan på olika ekvationer för att visa var varje strategi fungerar bäst.
1. Metod 1: Faktorisering — Lös x² − 7x + 12 = 0
Leta efter två heltal vars produkt är lika med c (här, 12) och vars summa är lika med b (här, −7). Heltalsparen som multipliceras till 12: 1 × 12, 2 × 6, 3 × 4, och deras negativa. Bland dessa multipliceras −3 och −4 till +12 och adderar till −7. Så x² − 7x + 12 = (x − 3)(x − 4) = 0. Enligt nollproduktegenskapen är antingen x − 3 = 0 eller x − 4 = 0, vilket ger x = 3 eller x = 4. Verifikation: (3)² − 7(3) + 12 = 9 − 21 + 12 = 0 ✓. (4)² − 7(4) + 12 = 16 − 28 + 12 = 0 ✓. Faktorisering är det snabbaste valet här — hela problemet tar cirka 20 sekunder när du ser faktorparet. Prova att faktorisera först när alla koefficienter är små heltal. Om du inte kan hitta heltalsfaktorer inom 10–15 sekunder, växla till den kvadratiska formeln istället för att tvinga det.
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.
Innan du löser, beräkna D = b² − 4ac. Om D < 0, finns det inga reella lösningar — klar. Om D ≥ 0, välj din metod: faktorisera om ett par visas inom 15 sekunder, använd formeln annars. För vertexform eller jämnt b med a = 1, slutför kvadraten.
Solving Radical Equations Step by Step
Kärnproceduren för att lösa en radikal ekvation har fyra steg: isolera radikalen på ena sidan, höj båda sidorna till den potens som motsvarar index, lös den resulterande ekvationen och kontrollera sedan varje kandidatlösning i originalet. Kontrollsteget är inte valfritt — främmande lösningar är vanliga i examensproblem och kan inte upptäckas på något annat sätt. Nedan demonstreras den fullständiga proceduren på fyra exempel av ökande komplexitet: en enkel kvadratrotsekvation, en kvadratrotsekvation där den resulterande ekvationen är linjär, en kubrotsekvation och en ekvation med två radikaltermer på samma sida.
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
Det mest testade scenariot på Algebra 2-proven är en radikal ekvation där höger sida är ett linjärt eller kvadratiskt uttryck i x. Efter kvadrering får du en kvadratisk ekvation som sedan måste lösas, och båda rötterna måste kontrolleras för främmande lösningar. Det är här lösningen av kvadratiska och radikala ekvationer överlappar direkt. Tre helt utarbetade exempel nedan täcker de tre huvudformerna: radikal lika med en linjär monom (√ = x), radikal lika med ett binomial (√ = x + n), och ett fall där radikanden själv innehåller x².
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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