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TMUA 2020 · Paper 1 · Question 9 of 20

TMUA 2020 Paper 1 Question 9

Algebra and functions — Roots and surds · rebuilding a quadratic. Try it first; the answer and a full worked solution are below.

TMUA 2020 · Paper 1Algebra and functionsRoots and surds · rebuilding a quadratic6 options
The quadratic expression x2 − 14x + 9 factorises as (x − α)(x − β), where α and β are positive real numbers.

Which quadratic expression can be factorised as (x − √α)(x − √β)?

  1. Ax2 − √10 x + 3
  2. Bx2 − √14 x + 3
  3. Cx2 − √20 x + 3
  4. Dx2 − 178x + 81
  5. Ex2 − 176x + 81
  6. Fx2 + 196x + 81
Show the answer and worked solution
answer · C
  1. Ax2 − √10 x + 3
  2. Bx2 − √14 x + 3
  3. Cx2 − √20 x + 3
  4. Dx2 − 178x + 81
  5. Ex2 − 176x + 81
  6. Fx2 + 196x + 81
There is no need to find α and β separately: α + β = 14 and αβ = 9. The new quadratic is x2 − (√α + √β)x + √αβ. The constant term is √9 = 3. For the middle coefficient, square it: (√α + √β)2 = α + β + 2√αβ = 14 + 6 = 20, so √α + √β = √20. The expression is x2 − √20 x + 3; option B is what you get by forgetting the 2√αβ cross term.