The Student Room Group
2x^2 - 8x - 11 = 0

deal with the first two terms only

2x^2 - 8x = 0

factorise out the 2

2(x^2 - 4x) = 0

square the bracket

2((x - 2)^2) = 0

subtract the number in the bracket and square it

2((x - 2)^2 - 2^2) = 0

bring down the constant

(x - 2)^2 - 8 - 11 = 0

(x - 2)^2 - 19 = 0
LazyWorseThanInfidel
2x^2 - 8x - 11 = 0

deal with the first two terms only

2x^2 - 8x = 0

factorise out the 2

2(x^2 - 4x) = 0

square the bracket

2((x - 2)^2) = 0

subtract the number in the bracket and square it

2((x - 2)^2 - 2^2) = 0

bring down the constant

(x - 2)^2 - 8 - 11 = 0

(x - 2)^2 - 19 = 0

All of these bolded lines are wrong. I can see what you're doing, and it's the right method, but please don't mislead people into thinking that these things are actually equal to 0. What you should be doing is:

2x^2 - 8x - 11 = 0
(2x^2 - 8x) - 11 = 0
2(x^2 - 4x) - 11 = 0
2[(x - 2)^2 - 2^2] - 11 = 0

...and for some reason, in the two underlined lines, you dropped the factor of 2.
sorry il have to step it up a league well spotted

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