bigfernhead wrote:

The way I did it was wrong, unfortunately.

I squared both sides right away (I somehow remembered learning this method a LONG time ago *sigh*), the equation is true for any value of X. So for me, A was sufficient. It's going to be hard to unlearn it.

the OA is B.

goldeneagle94 wrote:

bigfernhead wrote:

Is sqrt [(x-3)^2] = 3-x ?

1) x is not equal to 3

2) -x*lxl > 0

why am I not allowed to take the Square of both the left and the right side of the equation?

Then it would equal :

(x-3)^2 = (3-x)^2

If I do this, then A is true. I'm just not sure why you can't. (this is how I learned it back in HS).

So,

Can you explain how you solved this question ?

Old habits don't go away so easy.

I have the same issue as yours.

So, to overcome that, I have started taking the inequalities as is and first using the Substitution method.

For e.g.

In this problem, I would keep the question as is:

\(sqrt [(x-3)^2] = (3-x)\)

1) Substitute different values (e.g. x = -4, -1, 0, 1, 4) and check for its equality.

2) Subsitute different values in the statement itself to realize that x < 0. Then check the question for its equality with x = -4, -1,

This approach of NOT taking variables from LHS (left hand side of eqn) to RHS or vice-versa has been helping me.

If substitution doesn't work, then I go for shifting the variables from LHS or RHS to the other side.

Hope it helps you.