Algebra — Equations, Inequalities and Modulus

For ax² + bx + c = 0 with a ≠ 0, everything CAT asks follows from three quantities: the discriminant D = b² − 4ac, the sum of roots −b/a and the product of roots c/a. You almost never need the roots themselves.

• The sign of the roots, without solving. Product c/a > 0 means both roots carry the same sign; then the sum −b/a tells you which. Product c/a < 0 means the roots have opposite signs. Example: x² − 5x + 6 = 0 has product +6 and sum +5, so both roots are positive (they are 2 and 3). And x² − x − 6 = 0 has product −6, so the roots straddle zero (they are 3 and −2).
• Maximum and minimum. The parabola y = ax² + bx + c turns at x = −b/2a, and the turning value is c − b²/4a. If a > 0 it opens upward, so that value is the minimum; if a < 0 it opens downward and the value is the maximum. Example: the maximum of 6 + 4x − x² is at x = −4/(2 × −1) = 2, and equals 6 + 8 − 4 = 10.
• Building the equation back. If you are told the sum S and product P of the roots, the equation is x² − Sx + P = 0. This is how "the roots of one equation are twice those of another" questions are meant to be done: new sum = 2S, new product = 4P.
• Common roots. If ax² + bx + c = 0 and px² + qx + r = 0 share both roots, then a/p = b/q = c/r. If they share exactly one root, subtract the two equations — the x² terms cancel and the linear equation that remains gives the common root directly.
• Worked example. For which k does x² + kx + 9 = 0 have equal roots? Equal roots means D = 0, so k² − 4(1)(9) = 0, k² = 36, k = ±6. Note that both signs qualify — a question that expects only +6 has been written carelessly, and in a TITA question you would have to decide which the phrasing wants.

• Remainder theorem. The remainder when f(x) is divided by (x − a) is simply f(a). Nothing needs to be divided. Example: the remainder of x³ − 4x + 7 on division by (x − 2) is f(2) = 8 − 8 + 7 = 7.
• Factor theorem. (x − a) divides f(x) exactly when f(a) = 0. So "find k if (x − 3) is a factor of x³ − kx + 6" is one substitution: 27 − 3k + 6 = 0, k = 11.
• Where to look for integer roots. If a polynomial with integer coefficients and leading coefficient 1 has an integer root, that root divides the constant term. For x³ − 6x² + 11x − 6 the candidates are the divisors of 6: ±1, ±2, ±3, ±6. Testing 1 gives 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor, and the rest factors as (x − 2)(x − 3). Roots: 1, 2, 3.
• Sum and product for a cubic. For ax³ + bx² + cx + d = 0 the three roots satisfy: sum = −b/a, sum of products in pairs = c/a, product = −d/a. CAT uses these the same way it uses the quadratic versions — to answer without solving.

One method solves almost every polynomial and rational inequality CAT sets. It is worth learning as a fixed procedure, because the mistakes candidates make here are procedural rather than conceptual.

1. Move everything to one side so the inequality reads expression > 0 (or < 0, ≥ 0, ≤ 0). Never cross-multiply by something that could be negative — that is the single most common error in this topic.
2. Factor the expression completely and mark every root of the numerator and every root of the denominator on a number line.
3. The expression changes sign at each root of odd multiplicity and does not change sign at a root of even multiplicity (a squared factor).
4. Find the sign of the whole expression in the rightmost interval by plugging in a large number, then alternate leftward according to rule 3.
5. Include roots of the numerator when the inequality is ≥ or ≤. Always exclude roots of the denominator, whatever the sign, because the expression is undefined there.

• Worked example 1. Solve (x − 1)(x − 4) < 0. Roots at 1 and 4; for large x the product is positive, so signs run + − + from the right. The expression is negative between the roots: 1 < x < 4.
• Worked example 2. Solve (x − 2)/(x + 3) ≥ 0. Numerator root 2 (included), denominator root −3 (always excluded). For large x the ratio is positive. Signs from the right: + on x > 2, − on −3 < x < 2, + on x < −3. Answer: x < −3 or x ≥ 2. Cross-multiplying to get x − 2 ≥ 0 would have lost the entire left-hand branch.
• Worked example 3 (even multiplicity). Solve (x − 1)²(x − 5) > 0. The squared factor never changes the sign, so the expression is positive exactly where (x − 5) is positive, except at x = 1 where it is zero. Answer: x > 5.

|x| is the distance of x from zero, so it is never negative. Reading it as a distance rather than as "remove the sign" makes most questions transparent: |x − 3| is the distance between x and 3, and |x − 3| < 5 therefore means x lies within 5 of 3, that is −2 < x < 8. No cases needed.

• Two moduli: split at the critical points. To solve |x − 1| + |x − 4| = 5, mark the points where each modulus changes sign — 1 and 4 — and handle three regions. For x < 1: (1 − x) + (4 − x) = 5 gives x = 0 ✓ (0 < 1, so valid). For 1 ≤ x ≤ 4: (x − 1) + (4 − x) = 3 ≠ 5, no solution. For x > 4: (x − 1) + (x − 4) = 5 gives x = 5 ✓. Answer: x = 0 or x = 5. The distance reading confirms it: we want points whose distances to 1 and 4 total 5, and since the two are 3 apart, the point must lie 1 unit outside on either side.
• Always check candidates against the region you assumed. Squaring or case-splitting produces extra values that satisfy the algebra but not the original equation. Squaring |x − 2| = 2x − 7 gives x² − 4x + 4 = 4x² − 28x + 49, that is x² − 8x + 15 = 0, whose roots are 3 and 5. Test both in the original: at x = 3 the left side is |1| = 1 and the right side is 6 − 7 = −1 — rejected, because a modulus cannot be negative. At x = 5 the left side is |3| = 3 and the right side is 10 − 7 = 3 ✓. So the equation has exactly one solution, x = 5, and squaring manufactured the other. A candidate who skipped the check would answer "two solutions".
• Minimum of a sum of distances. |x − a| + |x − b| has minimum |a − b|, attained anywhere between a and b. For three points it is minimised at the middle one. This turns a hard-looking question into a one-line answer: the minimum of |x − 2| + |x − 6| + |x − 9| is at x = 6, and equals 4 + 0 + 3 = 7.

• TITA changes the arithmetic of guessing. A type-in question carries no negative marking, so any answer you can produce is free to submit. On an MCQ, a wrong answer costs 1 mark against a possible 3 — so a blind guess among four options has expected value (3 − 3)/4 = 0, and a guess you have narrowed to two options has positive expected value. Eliminate, then guess; never leave a TITA blank if you have any number at all.
• Substitute the options. When an MCQ gives four numbers and the algebra looks long, testing the options is often faster than solving — and it cannot go wrong, because the answer is definitely one of them. This is a legitimate method, not a shortcut to be ashamed of.
• Sanity-check with one number. Before trusting an algebraic manipulation, put in a convenient value — x = 1, or a = b = 1 — on both sides. A sign slip shows up immediately. Thirty seconds spent here saves a whole question.