Senior algebra problems (ages 16 to 18)

Answer: 14

1. Let the roots be α, β, γ. From x³ − 4x² + x + 6 (Vieta): α + β + γ = 4 and αβ + βγ + γα = 1.
2. Use (α + β + γ)² = α² + β² + γ² + 2(αβ + βγ + γα).
3. So α² + β² + γ² = 4² − 2 × 1 = 14.
4. Check: the cubic factorises as (x + 1)(x − 2)(x − 3), and 1 + 4 + 9 = 14. ✓ Answer 14.

Why this works: Symmetric expressions in the roots can be read straight from the coefficients (Vieta’s formulas) without solving the equation.

Where it leads: Newton’s identities extend this to any power sum of the roots without solving the cubic.

Answer: C, x = 64

1. Change every logarithm to base 2: log₄ x = (log₂ x)/2 and log₈ x = (log₂ x)/3.
2. Let L = log₂ x. Then L + L/2 + L/3 = 11.
3. L(6 + 3 + 2)/6 = 11L/6 = 11, so L = 6.
4. x = 2⁶ = 64.
5. Check: log₂ 64 + log₄ 64 + log₈ 64 = 6 + 3 + 2 = 11. ✓ x = 64 (C).

Why this works: Change of base, log_b^k x = (log_b x)/k, puts every term in the same unknown.

Where it leads: log_b x = log x / log b: changing base turns any mix of logarithms into multiples of one.

Answer: 25

1. x² + y² is the squared distance from the origin to the point (x, y), which lies on the line 3x + 4y = 25.
2. The shortest distance from the origin to the line ax + by = c is |c|/√(a² + b²) = 25/√(9 + 16) = 5.
3. It is reached at the foot of the perpendicular, (3, 4): indeed 3 × 3 + 4 × 4 = 25.
4. So the minimum of x² + y² is 5² = 25.
5. (Cauchy–Schwarz gives the same: 25² = (3x + 4y)² ≤ (9 + 16)(x² + y²).) Answer 25.

Why this works: A quadratic expression like x² + y² often has a geometric meaning; minimising distance to a line is the perpendicular.

Where it leads: Cauchy–Schwarz gives it in one line: 25 = 3x + 4y ≤ 5√(x² + y²).

Answer: A, −2

1. Put x = 2: f(2) + 2f(−1) = 3 × 4 = 12.
2. Put x = −1 (so that 1 − x = 2): f(−1) + 2f(2) = 3.
3. From the second equation, f(−1) = 3 − 2f(2).
4. Substitute into the first: f(2) + 6 − 4f(2) = 12, so −3f(2) = 6.
5. f(2) = −2 (A). (Then f(−1) = 7; check: −2 + 14 = 12 ✓.)

Why this works: Swapping x with 1 − x gives a second equation in the same two unknowns, so a pair of simultaneous equations appears.

Where it leads: The substitution x → 1 − x is an involution (doing it twice gets you back), which is why two equations close up.

Answer: C, 6

1. Let the first term be a and the ratio r, with |r| < 1. Then a/(1 − r) = 12.
2. Squaring each term gives a geometric series with first term a² and ratio r²: a²/(1 − r²) = 48.
3. Divide, using 1 − r² = (1 − r)(1 + r): [a²/((1 − r)(1 + r))] ÷ [a/(1 − r)] = a/(1 + r) = 48/12 = 4.
4. So a = 12(1 − r) and a = 4(1 + r): 12 − 12r = 4 + 4r, giving r = 1/2 and a = 6.
5. Check: 6 + 3 + 1.5 + … = 12 and 36 + 9 + 2.25 + … = 48. ✓ The first term is 6 (C).

Why this works: Factorising 1 − r² = (1 − r)(1 + r) makes the ratio of the two sums simple. Always check |r| < 1 so both series converge.

Where it leads: Squaring each term of a geometric series gives another geometric series with ratio r².

Answer: C, 3

1. Any solution has |x| = |3 sin x| ≤ 3, so all solutions lie in −3 ≤ x ≤ 3.
2. x = 0 is a solution.
3. For 0 < x ≤ 3 look at g(x) = 3 sin x − x: g(0) = 0 and g′(0) = 3 − 1 = 2 > 0, so g is positive just after 0.
4. g(3) = 3 sin 3 − 3 ≈ 0.42 − 3 < 0, so g crosses zero somewhere in (0, 3). On (0, 3), g″(x) = −3 sin x < 0, so g is concave and can cross zero only once there.
5. g is odd (g(−x) = −g(x)), so there is exactly one negative solution too.
6. Total: 3 solutions (C) (x = 0 and x ≈ ±2.28).

Why this works: Bounding the region (|x| ≤ 3) and using symmetry and concavity turns a transcendental equation into a picture you can trust.

Where it leads: By symmetry the solutions come in ± pairs, plus x = 0: an odd number of solutions.

Answer: 111

1. First differences: 7 − 3 = 4 and 13 − 7 = 6. Second difference: 2.
2. For P(x) = ax² + bx + c the second difference is 2a, so a = 1.
3. Then P(1) = 1 + b + c = 3 and P(2) = 4 + 2b + c = 7. Subtracting: 3 + b = 4, so b = 1, and c = 1.
4. P(x) = x² + x + 1. Check P(3) = 13. ✓
5. P(10) = 100 + 10 + 1 = 111.

Why this works: Finite differences identify polynomials: a quadratic has constant second differences equal to twice its leading coefficient.

Where it leads: A polynomial of degree n has constant n-th differences, which is why three values fix a quadratic.

Answer: 19/9

1. Vieta: α + β = 5, αβ = 3.
2. α² + β² = (α + β)² − 2αβ = 25 − 6 = 19.
3. 1/α² + 1/β² = 19/3² = 19/9.

Why this works: Symmetric expressions in the roots can be written using the sum and product, which Vieta gives straight from the coefficients.

Where it leads: 1/α and 1/β are the roots of the ‘reversed’ quadratic 3x² − 5x + 1 = 0: reversing coefficients inverts the roots.

Answer: 1023

1. x¹⁰ = (x − 1)(x − 2)Q(x) + ax + b for some polynomial Q.
2. x = 1: 1 = a + b. x = 2: 1024 = 2a + b.
3. Subtracting: a = 1023 (and b = −1022).

Why this works: Substituting the roots of the divisor kills the quotient term and leaves simple equations for the remainder.

Where it leads: This is polynomial interpolation: the remainder is the line through (1, 1) and (2, 1024). Dividing by a cubic would give the quadratic through three points.

Answer: C, 3/4

1. S = 1/3 + 2/9 + 3/27 + … and S/3 = 1/9 + 2/27 + 3/81 + …
2. Subtract term by term: S − S/3 = 1/3 + 1/9 + 1/27 + … = (1/3)/(1 − 1/3) = 1/2.
3. So (2/3)S = 1/2 and S = 3/4 (C).

Why this works: Shifting and subtracting turns an arithmetic-geometric series into a plain geometric one.

Where it leads: In general Σ n xⁿ = x/(1 − x)² for |x| < 1, which you can also get by differentiating the geometric series. Here x = 1/3 gives 3/4.

Answer: B, 64 + 2√2

1. Let t = log₈ x. Then log_x 8 = 1/t, so t + 1/t = 5/2, i.e. 2t² − 5t + 2 = 0.
2. t = 2 or t = 1/2, so x = 8² = 64 or x = 8^1/2 = 2√2.
3. Both are valid bases (positive, not 1). Sum: 64 + 2√2 (B).

Why this works: The two logs are reciprocals, so a substitution gives a reciprocal equation t + 1/t = k.

Where it leads: t + 1/t ≥ 2 for t > 0 (AM–GM), so log_x 8 + log₈ x = k has solutions with x > 1 only when k ≥ 2.

Answer: 4

1. x² + 4/x² = (x − 2/x)² + 4.
2. A square is at least 0, so the expression is at least 4.
3. Equality when x = 2/x, i.e. x² = 2. The minimum is 4.

Why this works: Writing the expression as a square plus a constant proves the bound and shows when it is reached.

Where it leads: This is AM–GM: a + b ≥ 2√(ab) with a = x², b = 4/x². Many minimisation problems need no calculus at all.

Answer: 35

1. With y = 1: f(x + 1) = f(x) + f(1) + 2x = f(x) + 2x + 3.
2. f(2) = 3 + 5 = 8, f(3) = 8 + 7 = 15, f(4) = 15 + 9 = 24, f(5) = 24 + 11 = 35.
3. f(5) = 35. (In fact f(x) = x² + 2x fits.)

Why this works: Substituting a simple value (y = 1) turns the functional equation into a recurrence that determines every value from f(1).

Where it leads: g(x) = f(x) − x² satisfies Cauchy’s equation g(x + y) = g(x) + g(y), whose solutions on the whole numbers are g(x) = cx.

Answer: −560

1. The x³ term comes from choosing −x three times and 2 four times.
2. Coefficient: C(7, 3) × 2⁴ × (−1)³ = 35 × 16 × (−1).
3. = −560.

Why this works: Each term of the binomial expansion records how many times each part of the bracket was chosen.

Where it leads: Putting x = 1 gives the sum of all coefficients: (2 − 1)⁷ = 1, a quick check on any expansion.

Answer: 3080

1. k(k + 1) = ⅓[k(k + 1)(k + 2) − (k − 1)k(k + 1)].
2. Summing from k = 1 to 20, the terms telescope, leaving ⅓ × 20 × 21 × 22.
3. = 9240/3 = 3080.

Why this works: Writing each term as a difference of consecutive ‘rising products’ makes the sum telescope, just like sums of 1/(k(k + 1)).

Where it leads: In general Σ k(k + 1)…(k + m − 1) = n(n + 1)…(n + m)/(m + 1): a discrete version of integrating x^m.