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CAT QuantmediumAlgebra

Algebra: Equations, Inequalities, Functions and Logs

Updated 22 October 2026

Five questions on quadratic equations, inequalities, functional equations and logarithms.

Question 1 · Multiple choice (+3 / −1)

If α and β are the roots of x² − 6x + 4 = 0, what is the value of α²/β + β²/α?

Show the solution

Answer: B. 36

  1. From the equation, α + β = 6 and αβ = 4.
  2. α²/β + β²/α = (α³ + β³)/αβ.
  3. α³ + β³ = (α + β)³ − 3αβ(α + β) = 216 − 3 × 4 × 6 = 216 − 72 = 144.
  4. So the value is 144 ÷ 4 = 36.
  5. Using (α + β)³ = 216 in place of α³ + β³ gives 54, a common slip.
Question 2 · Type in the answer (+3 / 0)

How many integer values of x satisfy |x − 3| + |x + 2| ≤ 11? Enter the number of values.

Show the solution

Answer: 12

  1. The critical points are x = −2 and x = 3.
  2. For −2 ≤ x ≤ 3, the sum is (3 − x) + (x + 2) = 5 ≤ 11, so every x here works.
  3. For x > 3: (x − 3) + (x + 2) = 2x − 1 ≤ 11, so x ≤ 6.
  4. For x < −2: (3 − x) − (x + 2) = 1 − 2x ≤ 11, so x ≥ −5.
  5. The solution is −5 ≤ x ≤ 6, which contains 6 − (−5) + 1 = 12 integers.
  6. Shortcut: |x − a| + |x − b| ≤ k is an interval centred at (a + b)/2 = 0.5 with half-width k/2 = 5.5, i.e. [−5, 6].
Question 3 · Multiple choice (+3 / −1)

What is the product of all real values of x satisfying x^(log₁₀ x) = 1000x², where x > 0?

Show the solution

Answer: B. 100

  1. Take log base 10 of both sides: (log x)(log x) = log 1000 + 2 log x.
  2. Let t = log x: t² = 3 + 2t, i.e. t² − 2t − 3 = 0, so (t − 3)(t + 1) = 0.
  3. t = 3 gives x = 1,000; t = −1 gives x = 0.1.
  4. Product = 1,000 × 0.1 = 100.
  5. Shortcut: the product of the x-values is 10^(sum of the t-roots) = 10² = 100.
Question 4 · Type in the answer (+3 / 0)

A function f satisfies f(x) + 2f(1/x) = 3x for every non-zero real x. What is the value of f(1/2)? Enter the answer as a decimal.

Show the solution

Answer: 3.5

  1. Put x = 2: f(2) + 2f(1/2) = 6.
  2. Put x = 1/2: f(1/2) + 2f(2) = 3/2.
  3. From the first, f(2) = 6 − 2f(1/2). Substitute: f(1/2) + 12 − 4f(1/2) = 3/2.
  4. So −3f(1/2) = 3/2 − 12 = −21/2, giving f(1/2) = 7/2 = 3.5.
  5. In general f(x) = 2/x − x, which gives f(1/2) = 4 − 0.5 = 3.5.
Question 5 · Multiple choice (+3 / −1)

For how many integers k with −20 ≤ k ≤ 20 does the equation x² + kx + k + 3 = 0 have two distinct real roots?

Show the solution

Answer: C. 32

  1. Two distinct real roots need a positive discriminant: k² − 4(k + 3) > 0.
  2. k² − 4k − 12 > 0 factorises as (k − 6)(k + 2) > 0, so k > 6 or k < −2.
  3. k = 7 to 20 gives 14 values; k = −20 to −3 gives 18 values.
  4. Total = 14 + 18 = 32.
  5. Shortcut: 41 integers in all, minus the 9 values from −2 to 6 where the discriminant is not positive. Including k = −2 and 6 (D = 0) gives 34.

Full worked solution

  1. For ax² + bx + c = 0: sum of roots = −b/a, product = c/a; express symmetric functions through them, e.g. α³ + β³ = (α + β)³ − 3αβ(α + β).
  2. Distinct real roots need D = b² − 4ac > 0; equal roots need D = 0.
  3. Modulus inequalities: split at the critical points, or use the geometric meaning of |x − a| as distance from a.
  4. Equations with x in the exponent: take logs and substitute t = log x.
  5. Functional equations in f(x) and f(1/x): substitute x and 1/x to get two linear equations and solve.