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Note that in September I am due to start as a trainee maths teacher . For my interview, I taught a lesson on inequalities to a class of GCSE pupils. In particular, we focused on converting worded problems...

(x² + 6x + 9) - 49 = 0
(x + 3)2 - 49 = 0

then to solve

(x + 3)² = 49
x + 3 = +-7
x = -10 or x = 4

2x² + 3x = 92x² + 3x - 9 = 0(2x - 3)(x+3)2x-3 = 02x = 3x = 3/2x+3=0x=-3x = 3/2 or -3

The first thing we notice is that there is a common factor of x. Thus we can factor it out and get:x(x^2 - 10x + 16)
This becomes much easier to deal with. We now focus on the equation in the bracket...

Split up the two terms: cos30° = (1/2)√3 and tan60° = √3 (from memorised values of cos and tan)Therefore 16cos30° = 8√3 and 5tan60° = 5√3 8√3 - 5√3 = 3√3 = √9 x √3 = √27x=27