Set of values?

Question

Find the set of values of x for which (x - 6)^2 > 0

Ans: x < 4 or x > 9

Please show your workings clearly and explain..thanks

Answer 1

I think you have transcribed the problem incorrectly because all squares will be greater than zero... I think you meant to write:

(x-6)² > x

You start by multiplying out the (x-6)(x-6) on the left:
x² - 12x + 36 > x

Now subtract x from both sides:
x² - 13x + 36 > 0

Now factor:
(x - 4)(x - 9) > 0

Now for this to work, either both factors (x-4) and (x-9) are positive (+ times + = +) or both are negative (- times - = +)

x - 4 and x - 9 will only both be positive if x > 9.
x - 4 and x - 9 will only both be negative if x < 4.

So there is your answer:
x < 4 or x > 9

Answer 2

your answers are wrong. any value for x that is real and is not 6 will satisfy this equation. for example is you were to put 5 in the equation you have there (5 - 6)^2 = 1 which is greater than 0.

Answer 3

(x-6)^2 > 0

1st of all everything (including negative values) squared is positive (try it on your calculator). Only "0" is "0" if squared! So all we need to do, is find out, when your brackets are "0".

Therefor: (x-6)=0 --> x=6

So only in case x=6 you will not be greater then "0"!!!

Answer 4

Your answer is wrong
The correct answer is x=/= 6