Hi,
|a| = a-3.
As |a| is always >= 0, a>=3.
a can only take values greater than or equal to 3.
So, by squaring both sides, if you are getting a value for 'a'<3, then it should not be a solution. As a=3/2, which contradicts a>=3, there should be no solution.

I will give you an analogy
What is the solution for |x| = -2. We know |x| cannot be negative. So, straight away we can say no solution.
But squaring we get x^2 = 4 => x=-2 or +2.
Whenever we square both sides of equations involving modulus , first make sure the quantity equal to modulus is non-negative.

thanks for responding!

a can only take values greater than or equal to 3.

I dont agree with that. I thought about that already.

Let's take a = 4 ; |a-3| = 1 != 4. this equation doesnt have any solution. However, when you square it, it does have a solution!

I am not able to understand what's wrong.......

voodoo_child wrote:thanks for responding!

a can only take values greater than or equal to 3.

I dont agree with that. I thought about that already.

Hey,
I don't think you understood my point. Probably, my post wasn't too clear. I will try to elaborate it.
|a| = a-3 is given
LHS is >= 0 right?
So, RHS >= 0
So, a-3>=0. So, for finding the solution for this, you can only consider a>=3 as a<3 is definitely not going to give any solution.
So, for a<3 there is no solution
Now for a>=3, we solve this by your squaring method and we get a=3/2. But, this doesn't satisfy a>=3.
So, even for a>=3, we don't get any solution.

Let's take a = 4 ; |a-3| = 1 != 4. this equation doesnt have any solution. However, when you square it, it does have a solution!

I am not able to understand what's wrong.......

Why does |a-3| = 1 not have solution? Any typo?
Anyway, this is not an illustration for the case we are dealing with. |a-3| = -1 would be an apt example.

let me try another route

ques: |a|= a-3

For a>0 |a| = a
so a = a-3 or 0 = -3; No Solution

For a<0 |a| = -a
so -a = a-3 or 2a = 3 or a = 3/2
but a has to be negative, so there is no solution

As for squaring both sides, the basic condition that needs to be satisfied is that a-3 > 0 (because a-3 is equal to a mod function which is always positive), so a > 3
Solving through squaring gives a = 3/2 which is not >3 so no solution.

Hope this helps.