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TMUA 2017 · Paper 1 · Question 19 of 20

TMUA 2017 Paper 1 Question 19

Algebra and functions — Quadratic inequalities · scaling the roots. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 1Algebra and functionsQuadratic inequalities · scaling the roots6 optionshard
The set of solutions to the inequality x2 + bx + c < 0 is the interval p < x < q where b, c, p and q are real constants with c < 0.

In terms of p, q and c, what is the set of solutions to the inequality x2 + bcx + c3 < 0?

  1. Apc < x < qc
  2. Bqc < x < pc
  3. Cpc < x < qc
  4. Dqc < x < pc
  5. Epc2 < x < qc2
  6. Fqc2 < x < pc2
Show the answer and worked solution
answer · D
  1. Apc < x < qc
  2. Bqc < x < pc
  3. Cpc < x < qc
  4. Dqc < x < pc
  5. Epc2 < x < qc2
  6. Fqc2 < x < pc2
The roots of x2 + bx + c = 0 are p and q. Test x = pc in the new quadratic: (pc)2 + bc(pc) + c3 = c2(p2 + bp + c) = 0, and the same works for qc, so the new roots are exactly the old ones multiplied by c. The new quadratic still opens upwards, so its solution set is the interval between its roots. Since c < 0, multiplying by c reverses the order: p < q becomes qc < pc, so the set is qc < x < pc.