check you quadratic equation for (1) again.....

you have x-x^2=2

then it should follow:

x-x^2-2=0

-x^2+x-2=0

x^2-x+2=0 (you can use the quadratice equation formula)

you get x=2 or x=-1 (which you do, I am just making sure people understand what you did)

now just do what you would with (2) and you get the same factors right...

go ahead try plugging in (-1) for x in (2), it will not add upto 2!

only (2) works for x.....

you were almost there, but you have plug in the numbers to verify...

Regan used to say, "trust but verify"

Antmavel wrote:

I would say E but i took me around 3mn

(1) |x-|x^2||=2

for each abs value you have to imagine it can be negative or positive before executing the abolute value. For example :

|x-|x^2||=2 ; x-x^2=2; x^2-x-2=0 ; (x-2)(x+1) so 2 answers : 2 and -1

x-|x^2|=2 ; no need to check this, first go to statement (2) to see if you find different numbers than in (1)

(2) |x^2-|x||=2

X^2-x=2 ; x^2-x-2=0 ; (x-2)(x+1), same answer than in statement 1

even if there are other possibilities, you've already found 2 different possible answers which are 2 and -1 in both statements so you can not choose any value for sure. E. no need to calculate everything.