vipulgoyal wrote:Hi Atekihcan, got the point but please explain how you are opening modes to get these values
This is possible only if x = -2 and x = 2.
I did not solve this problem by opening absolute value problem. My logic is as follows,
|x² - |x|| = 2 means the distance between |x| and x² is 2 and 0it is given that x is an integer.
As the expression involves x² and |x|, it does not matter whether x is negative or positive.
If some x = k satisfy the equation, x = -k will also satisfy.
Now, if x ≥ 3, the distance between x² and |x| will be always greater than 2.
And, if x = 0 or x = 1, x² = |x|
Only, possible solution is x = 2
So, x = -2 is another solution.
If you want to solve this problem by opening modulus brackets, here you go...
As x is an integer and x = 0 or x = 1 cannot be the solution of the equation, |x| must be greater than 1.
So, x² is always greater than |x|
So, |x² - |x|| = x² - |x| = 2
Now, if x > 1, x² - |x| = x² - x = 2 ---> x² - x - 2 = 0 ---> (x + 1)(x - 2) = 0 ---> x = 2
And, if x < -1, x² - |x| = x² - (-x) = 2 ---> x² + x - 2 = 0 ---> (x - 1)(x + 2) = 0 ---> x = -2
Aside 'mode' is very different thing (a statistical measure).
This problems are on absolute values or modulus. Some people refer it as 'mod' but that has another different interpretation in number theory.
