TMUA Practice Question — Worked Solution

P2·Algebra & Surds·5.5CrackTMUA Original

extraneous-rootssquaring-both-sidessurd-equation

Below is an attempted proof of the claim 'the equation

x + 4 = x - 2

has solutions

x = 0

and

x = 5

'. At which line does the argument first break down?

Line 1

Squaring both sides gives

x + 4 = (x - 2)^{2} = x^{2} - 4 x + 4

Line 2

This rearranges to

x^{2} - 5 x = 0

, i.e.

x (x - 5) = 0

Line 3

Therefore

x = 0

x = 5

Line 4

Hence the original equation has the two solutions

x = 0

and

x = 5

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