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TMUA 2018 · Paper 2 · Question 9 of 20

TMUA 2018 Paper 2 Question 9

Proof and counterexample — Finding the error · dividing by something that can be zero. Try it first; the answer and a full worked solution are below.

TMUA 2018 · Paper 2Proof and counterexampleFinding the error · dividing by something that can be zero7 options
Consider the following attempt to solve the equation 4x√2x−1 = 10x − 5:

4x√2x−1 = 10x − 5
(I)    4x√2x−1 = 5(2x−1)
(II)   16x2(2x−1) = 25(2x−1)2
(III)  16x2 = 25(2x−1)
(IV)   16x2 − 50x + 25 = 0
(V)    (8x−5)(2x−5) = 0
(VI)   The solutions of the original equation are x = 58 and x = 52.

Which one of the following is true?

  1. AThe solution is correct.
  2. BOnly one of x = 58 and x = 52 is correct and the error arises as a result of step (II).
  3. COnly one of x = 58 and x = 52 is correct and the error arises as a result of step (III).
  4. DOnly one of x = 58 and x = 52 is correct and the error arises as a result of step (IV).
  5. EThere is another value of x that satisfies the original equation and the error arises as a result of step (II).
  6. FThere is another value of x that satisfies the original equation and the error arises as a result of step (III).
  7. GThere is another value of x that satisfies the original equation and the error arises as a result of step (IV).
Show the answer and worked solution
answer · F
  1. AThe solution is correct.
  2. BOnly one of x = 58 and x = 52 is correct and the error arises as a result of step (II).
  3. COnly one of x = 58 and x = 52 is correct and the error arises as a result of step (III).
  4. DOnly one of x = 58 and x = 52 is correct and the error arises as a result of step (IV).
  5. EThere is another value of x that satisfies the original equation and the error arises as a result of step (II).
  6. FThere is another value of x that satisfies the original equation and the error arises as a result of step (III).
  7. GThere is another value of x that satisfies the original equation and the error arises as a result of step (IV).
Check the two answers first: at x = 58 both sides equal 54, and at x = 52 both sides equal 20, so neither is spurious. Squaring at step (II) is therefore harmless here. The damage is at step (III), which cancels a factor of 2x − 1 from both sides — legitimate only if 2x − 1 ≠ 0. At x = 12 both sides of the original equation are zero, so a third solution has been thrown away.