Equation solving sits inside the exam's largest category (Algebraic Expressions, Equations and Inequalities, 20%) and appears throughout the others. On the exam, wrong answer choices are built from specific solving errors — sign slips, skipped checks, dropped cases — so this lesson names each trap explicitly.
Distribute, collect variable terms on one side, constants on the other:
3(x − 4) = 5x + 2 → 3x − 12 = 5x + 2 → −14 = 2x → x = −7
Two errors account for nearly all linear-equation misses: failing to distribute across every term in the parentheses, and a sign flip when moving terms across the equals sign. Sanity check by substituting back: 3(−11) = −33 and 5(−7) + 2 = −33. ✓
| Method | Best when | Example |
|---|---|---|
| Factoring | integer roots exist | x² − 5x − 14 = 0 → (x − 7)(x + 2) = 0 → x = 7, −2 |
| Completing the square | coefficient of x is even; exact form wanted | see below |
| Quadratic formula | always works | x = [−b ± √(b² − 4ac)] / (2a) |
Try factoring for ten seconds; if it doesn't crack, go to the formula.
Completing the square, worked on x² − 6x + 4 = 0:
x² − 6x = −4
x² − 6x + 9 = −4 + 9 ← add (6/2)² = 9 to BOTH sides
(x − 3)² = 5
x − 3 = ±√5 → x = 3 ± √5
The binomial is (x − 3)², using half the x-coefficient — and whatever you add on the left must be added on the right.
For ax² + bx + c = 0 with real coefficients, the discriminant D = b² − 4ac determines the solution type:
| D | Solutions |
|---|---|
| D > 0 | two distinct real solutions |
| D = 0 | exactly one real solution (a repeated root) |
| D < 0 | two nonreal complex solutions — always a conjugate pair |
Example: 2x² − 3x + 5 = 0 has D = 9 − 40 = −31 < 0 → two nonreal complex conjugate solutions. No solving required, and the exam asks exactly this: "how many real solutions?"
When D < 0, the ± in the quadratic formula produces a ± on a square root of a negative number:
x² − 4x + 13 = 0: x = [4 ± √(16 − 52)]/2 = [4 ± √(−36)]/2 = [4 ± 6i]/2 = 2 ± 3i
Because the coefficients are real, nonreal solutions always come in conjugate pairs a ± bi: if 3 − 2i solves a quadratic with real coefficients, 3 + 2i solves it too. One real and one nonreal solution is impossible. Rebuilding the quadratic from a conjugate pair is fast with sum and product: roots 3 ± 2i have sum 6 and product 3² + 2² = 13, giving x² − 6x + 13 = 0 (x² − (sum)x + product).
√(x + 7) = x − 5
x + 7 = (x − 5)² = x² − 10x + 25
x² − 11x + 18 = 0 → (x − 2)(x − 9) = 0 → x = 2 or x = 9
Squaring both sides is not reversible — it can create solutions the original equation never had. Check both candidates in the original:
Solution: x = 9 only. The exam always lists the extraneous candidate among the choices; the check is where the point lives. Corollary worth memorizing: √(expression) = (negative number) has no solution — don't even square.
Values that zero any denominator are excluded before you start. Solve, then compare candidates to the exclusion list.
Simple case — cross-multiply when it's one fraction per side:
5/(x + 1) = 3/(x − 1) → 5(x − 1) = 3(x + 1) → 5x − 5 = 3x + 3 → x = 4 ✓ (4 ≠ −1, 1)
Trap case — the algebra lands exactly on an excluded value:
x/(x − 2) = 2/(x − 2) + 5 (x ≠ 2)
x = 2 + 5(x − 2) → x = 5x − 8 → x = 2 ✗ excluded
The only candidate is forbidden, so the equation has no solution. This is a deliberate exam design, not an edge case.
|X| = c means X is at distance c from 0:
| Form | Rewrite as | Notes |
|---|---|---|
| |X| = c, c > 0 | X = c or X = −c | two cases, always both |
| |X| = 0 | X = 0 | one solution |
| |X| = c, c < 0 | no solution | distance is never negative |
Worked: |2x − 3| = 7 → 2x − 3 = 7 or 2x − 3 = −7 → x = 5 or x = −2. And |x + 4| = −3 has no solution — stop at sight, no cases needed.
Q1. B. Distribute: 3x − 12 = 5x + 2; collect: −14 = 2x; so x = −7. Check: 3(−11) = −33 = 5(−7) + 2. ✓ A) flips the sign when dividing −14 by 2. C) fails to distribute the 3 to the −4 (3x − 4 = 5x + 2 gives x = −3). D) combines both errors. E) mis-collects, dividing −14 by 8 as if the variable terms added to 8x. Fix: distribute to every term inside the parentheses, then move terms with explicit sign changes — and substitute the answer back.
Q2. E. Need factors of −14 summing to −5: −7 and +2, so (x − 7)(x + 2) = 0 → x = 7, −2. A) reverses both signs — those are the roots of x² + 5x − 14. B) drops the sign on one root; check: 2 is not a root (4 − 10 − 14 ≠ 0). C) makes both roots negative, matching x² + 9x + 14 instead. D) uses the factor pair 14 and 1 with careless signs. Fix: the roots' signs are the opposites of the signs inside the factors — always substitute one root back to confirm.
Q3. C. x² − 6x = −4; add (6/2)² = 9 to both sides: (x − 3)² = 5, so x = 3 ± √5. A) writes the binomial as (x + 3)², flipping the shift — the sign in the binomial matches the sign pattern (x − 3)² for −6x. B) adds 9 to the left but effectively adds 13 on the right by mishandling the −4 (−4 + 9 = 5, not 13). D) uses the full coefficient 6 instead of half of it. E) forgets the square root, reporting (x − 3)² = 5 as x − 3 = ±5. Fix: half the x-coefficient goes in the binomial; its square goes on both sides; finish with a square root, not the raw constant.
Q4. D. D = (−3)² − 4(2)(5) = 9 − 40 = −31 < 0, so both solutions are nonreal — and with real coefficients they form a conjugate pair. A) requires D > 0. B) requires D = 0. C) is impossible for real coefficients: nonreal roots cannot appear alone, since the conjugate of a root is also a root. E) confuses "no real solutions" with "no solutions" — the complex solutions exist. Fix: compute b² − 4ac before solving; its sign alone answers every "how many real solutions" question.
Q5. A. x = [4 ± √(16 − 52)]/2 = [4 ± 6i]/2 = 2 ± 3i. B) flips the sign of the real part, using +b instead of −b in the formula. C) drops the i, treating √(−36) as √36. D) forgets to divide by 2a = 2. E) loses the real part entirely, as if the equation were x² + 9 = 0. Fix: simplify √(−n) as i√n first, then divide every term of the numerator by 2a.
Q6. E. Squaring gives x² − 11x + 18 = 0 → x = 2 or 9. Checking the original: x = 9 gives √16 = 4 = 9 − 5 ✓; x = 2 gives √9 = 3 ≠ −3 ✗. Only 9 survives. A) picks exactly the ghost — the candidate that fails the check. B) skips the check and keeps both candidates. C) includes −2, which solves nothing and isn't even a candidate. D) over-rejects: one candidate does verify. Fix: after squaring, substitute every candidate into the original equation; a principal square root can never equal a negative number.
Q7. D. Cross-multiply: 5(x − 1) = 3(x + 1) → 5x − 5 = 3x + 3 → 2x = 8 → x = 4; and 4 zeroes no denominator. ✓ A) swaps the multiplication, computing 5(x + 1) = 3(x − 1). B) reports an excluded value of the right-hand denominator as the answer. C) inverts the final division, computing 2/8. E) misapplies the no-solution trap — the candidate here is not excluded, so it stands. Fix: cross-multiply each numerator with the opposite denominator, then verify the solution against the excluded values (here x ≠ −1, 1).
Q8. A. Multiplying by (x − 2) gives x = 2 + 5(x − 2) → x = 2 — but x = 2 zeroes both original denominators, so it must be rejected, leaving no solution. B) accepts a candidate that makes the original expressions undefined. C) invents a second candidate; −2 never arises and doesn't satisfy the equation. D) confuses "no solution" with "identity" — substituting any other value (say x = 3: 3 vs. 2 + 5 = 7) shows the equation is false elsewhere. E) the algebra is actually correct; the flaw is only in accepting the candidate. Fix: list excluded values before solving a rational equation; any candidate on that list is rejected, and if none remain, the answer is "no solution."
Q9. B. |2x − 3| = 7 splits: 2x − 3 = 7 → x = 5, or 2x − 3 = −7 → x = −2. A) solves only the positive case. C) botches the negative case's sign: 2x = −4 gives −2, not 2. D) flips signs on both candidates. E) moves the −3 with the wrong sign in both cases (2x = 7 − 3 and 2x = −7 + 3), producing x = 2 and x = −2. Fix: |X| = c always generates two equations, X = c and X = −c — solve both completely before choosing an answer.
Q10. D. An absolute value is a distance and is never negative, so |x + 4| = −3 is impossible: no solution. A) and B) come from mechanically splitting into x + 4 = ±3 and solving anyway (x = −7, x = −1). C) keeps both of those invalid candidates. E) solves |x + 4| = 0 instead. Fix: before splitting any absolute value equation into cases, look at the right side — if it's negative, stop and answer "no solution."
Q11. E. Real coefficients force the conjugate 3 + 2i to be the other root. Sum = 6, product = (3)² + (2)² = 13, so x² − 6x + 13 = 0. Verify: (3 − 2i)² − 6(3 − 2i) + 13 = (9 − 12i − 4) − 18 + 12i + 13 = 0. ✓ A) flips the sign of the sum, matching roots −3 ± 2i. B) and C) make the product negative, but the product (a + bi)(a − bi) = a² + b² is always positive. D) uses product 3² − 2² = 5, computing a² − b² instead of a² + b². Fix: for conjugate roots a ± bi, the quadratic is x² − 2ax + (a² + b²) = 0 — sum flips sign, product adds the squares.
Q12. C. The left side, √(2x + 3), is a principal square root and is ≥ 0 wherever it is defined; it can never equal −5, so the original equation has no solution and x = 11 is a ghost created by squaring (check: √25 = 5 ≠ −5). A) squaring is not reversible — it merges the equation with √(2x + 3) = +5, whose solution is exactly the x = 11 found. B) satisfying the squared equation is not enough; candidates must satisfy the original. D) the squaring step itself was executed correctly; the error is accepting the candidate. E) the equation is false for every x in the domain, not true. Fix: if a principal square root is set equal to a negative number, declare "no solution" immediately — and in all other radical equations, check candidates in the original.
Q1. B. Distribute: 3x − 12 = 5x + 2; collect: −14 = 2x; so x = −7. Check: 3(−11) = −33 = 5(−7) + 2. ✓ A) flips the sign when dividing −14 by 2. C) fails to distribute the 3 to the −4 (3x − 4 = 5x + 2 gives x = −3). D) combines both errors. E) mis-collects, dividing −14 by 8 as if the variable terms added to 8x. Fix: distribute to every term inside the parentheses, then move terms with explicit sign changes — and substitute the answer back.
Q2. E. Need factors of −14 summing to −5: −7 and +2, so (x − 7)(x + 2) = 0 → x = 7, −2. A) reverses both signs — those are the roots of x² + 5x − 14. B) drops the sign on one root; check: 2 is not a root (4 − 10 − 14 ≠ 0). C) makes both roots negative, matching x² + 9x + 14 instead. D) uses the factor pair 14 and 1 with careless signs. Fix: the roots' signs are the opposites of the signs inside the factors — always substitute one root back to confirm.
Q3. C. x² − 6x = −4; add (6/2)² = 9 to both sides: (x − 3)² = 5, so x = 3 ± √5. A) writes the binomial as (x + 3)², flipping the shift — the sign in the binomial matches the sign pattern (x − 3)² for −6x. B) adds 9 to the left but effectively adds 13 on the right by mishandling the −4 (−4 + 9 = 5, not 13). D) uses the full coefficient 6 instead of half of it. E) forgets the square root, reporting (x − 3)² = 5 as x − 3 = ±5. Fix: half the x-coefficient goes in the binomial; its square goes on both sides; finish with a square root, not the raw constant.
Q4. D. D = (−3)² − 4(2)(5) = 9 − 40 = −31 < 0, so both solutions are nonreal — and with real coefficients they form a conjugate pair. A) requires D > 0. B) requires D = 0. C) is impossible for real coefficients: nonreal roots cannot appear alone, since the conjugate of a root is also a root. E) confuses "no real solutions" with "no solutions" — the complex solutions exist. Fix: compute b² − 4ac before solving; its sign alone answers every "how many real solutions" question.
Q5. A. x = [4 ± √(16 − 52)]/2 = [4 ± 6i]/2 = 2 ± 3i. B) flips the sign of the real part, using +b instead of −b in the formula. C) drops the i, treating √(−36) as √36. D) forgets to divide by 2a = 2. E) loses the real part entirely, as if the equation were x² + 9 = 0. Fix: simplify √(−n) as i√n first, then divide every term of the numerator by 2a.
Q6. E. Squaring gives x² − 11x + 18 = 0 → x = 2 or 9. Checking the original: x = 9 gives √16 = 4 = 9 − 5 ✓; x = 2 gives √9 = 3 ≠ −3 ✗. Only 9 survives. A) picks exactly the ghost — the candidate that fails the check. B) skips the check and keeps both candidates. C) includes −2, which solves nothing and isn't even a candidate. D) over-rejects: one candidate does verify. Fix: after squaring, substitute every candidate into the original equation; a principal square root can never equal a negative number.
Q7. D. Cross-multiply: 5(x − 1) = 3(x + 1) → 5x − 5 = 3x + 3 → 2x = 8 → x = 4; and 4 zeroes no denominator. ✓ A) swaps the multiplication, computing 5(x + 1) = 3(x − 1). B) reports an excluded value of the right-hand denominator as the answer. C) inverts the final division, computing 2/8. E) misapplies the no-solution trap — the candidate here is not excluded, so it stands. Fix: cross-multiply each numerator with the opposite denominator, then verify the solution against the excluded values (here x ≠ −1, 1).
Q8. A. Multiplying by (x − 2) gives x = 2 + 5(x − 2) → x = 2 — but x = 2 zeroes both original denominators, so it must be rejected, leaving no solution. B) accepts a candidate that makes the original expressions undefined. C) invents a second candidate; −2 never arises and doesn't satisfy the equation. D) confuses "no solution" with "identity" — substituting any other value (say x = 3: 3 vs. 2 + 5 = 7) shows the equation is false elsewhere. E) the algebra is actually correct; the flaw is only in accepting the candidate. Fix: list excluded values before solving a rational equation; any candidate on that list is rejected, and if none remain, the answer is "no solution."
Q9. B. |2x − 3| = 7 splits: 2x − 3 = 7 → x = 5, or 2x − 3 = −7 → x = −2. A) solves only the positive case. C) botches the negative case's sign: 2x = −4 gives −2, not 2. D) flips signs on both candidates. E) moves the −3 with the wrong sign in both cases (2x = 7 − 3 and 2x = −7 + 3), producing x = 2 and x = −2. Fix: |X| = c always generates two equations, X = c and X = −c — solve both completely before choosing an answer.
Q10. D. An absolute value is a distance and is never negative, so |x + 4| = −3 is impossible: no solution. A) and B) come from mechanically splitting into x + 4 = ±3 and solving anyway (x = −7, x = −1). C) keeps both of those invalid candidates. E) solves |x + 4| = 0 instead. Fix: before splitting any absolute value equation into cases, look at the right side — if it's negative, stop and answer "no solution."
Q11. E. Real coefficients force the conjugate 3 + 2i to be the other root. Sum = 6, product = (3)² + (2)² = 13, so x² − 6x + 13 = 0. Verify: (3 − 2i)² − 6(3 − 2i) + 13 = (9 − 12i − 4) − 18 + 12i + 13 = 0. ✓ A) flips the sign of the sum, matching roots −3 ± 2i. B) and C) make the product negative, but the product (a + bi)(a − bi) = a² + b² is always positive. D) uses product 3² − 2² = 5, computing a² − b² instead of a² + b². Fix: for conjugate roots a ± bi, the quadratic is x² − 2ax + (a² + b²) = 0 — sum flips sign, product adds the squares.
Q12. C. The left side, √(2x + 3), is a principal square root and is ≥ 0 wherever it is defined; it can never equal −5, so the original equation has no solution and x = 11 is a ghost created by squaring (check: √25 = 5 ≠ −5). A) squaring is not reversible — it merges the equation with √(2x + 3) = +5, whose solution is exactly the x = 11 found. B) satisfying the squared equation is not enough; candidates must satisfy the original. D) the squaring step itself was executed correctly; the error is accepting the candidate. E) the equation is false for every x in the domain, not true. Fix: if a principal square root is set equal to a negative number, declare "no solution" immediately — and in all other radical equations, check candidates in the original.