TTMUA Lab
TMUA 2019 · Paper 2 · Question 19 of 20

TMUA 2019 Paper 2 Question 19

Algebra and functions — Surds · recognising a perfect square. Try it first; the answer and a full worked solution are below.

TMUA 2019 · Paper 2Algebra and functionsSurds · recognising a perfect square7 options
Find the value of the expression √8 − 4√2 + 1 + √9 − 12√2 + 8
  1. A√26 − 16√2
  2. B4√2 − 4
  3. C−2
  4. D4 − 4√2
  5. E2
  6. F√26 − 4√2
  7. G1
Show the answer and worked solution
answer · E
  1. A√26 − 16√2
  2. B4√2 − 4
  3. C−2
  4. D4 − 4√2
  5. E2
  6. F√26 − 4√2
  7. G1
The unsimplified way each expression is written is the hint: each is already laid out as u2 − 2uv + v2. The first is (2√2)2 − 2(2√2)(1) + 12 = (2√2 − 1)2, and the second is 32 − 2(3)(2√2) + (2√2)2 = (3 − 2√2)2. A square root is the non-negative value, so watch the signs: 2√2 ≈ 2.83, so the first root is 2√2 − 1 and the second is 3 − 2√2. Adding, the surds cancel and the total is 2.