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Complements and parameters: sets of quadratic roots

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      Sets · Quadratic equations

      Question

      Let A = {x ∈ ℝ | x² − 3x + b = 0} and B = {x ∈ ℝ | (x − 2)(x² + 3x − 4) = 0}. If (ℝ \ B) ∩ A = ∅, find all real values of b.

      Venn diagram for complements and inclusionThe universal set ℝ contains the ellipse B{-4, 1, 2}. Its complement is outside the ellipse. The condition (ℝ \ B) ∩ A = ∅ is equivalent to A ⊆ B.Universal set ℝComplement ℝ \ BB = {-4, 1, 2}Element: −4A ⊆ BRoots: 1 and 2 (1+2=3)(∁_R B) ∩ A = ∅ ⟺ A ⊆ B
      The complement condition is equivalent to A ⊆ B. Either A is empty (b > 9/4), or A = {1, 2} (b = 2).

      Step-by-step solution

      1. Find B

      Factor (x − 2)(x² + 3x − 4) = (x − 2)(x + 4)(x − 1). Its real roots give B = {−4, 1, 2}.

      2. Rewrite the set condition

      A has no element outside B, so (ℝ \ B) ∩ A = ∅ is equivalent to A ⊆ B.

      3. Consider the number of real roots

      The discriminant is Δ = 9 − 4b. If Δ < 0, then b > 9/4 and A = ∅, which is valid. If Δ = 0, the repeated root is 3/2, which is not in B. If Δ > 0, two distinct elements of B must sum to 3 by Vieta's formulas. Only 1 and 2 do, giving b = 1 × 2 = 2; its discriminant is positive.

      4. Combine the valid cases

      The full answer is b = 2 or b > 9/4.

      The parameter values are b = 2 or b > 9/4.

      Which method does this problem use?

      Simplify the set relationship first. Then split by the number of real roots, including the empty set, and use the sum and product of roots to constrain the possible elements.