In how many ways can a pair of integers (x , a) be chosen such that $$x^{2}-2\mid x\mid+\mid a-2\mid=0$$ ?

$$x^{2}-2\mid x\mid+\mid a-2\mid=0$$

where x>= 0 and x>=2

$$x^2-2x+a-2\ =0$$ Using quadratic equation we have $$x=\ 1+\sqrt{\ 3-a}\ and\ x=1-\sqrt{\ 3-a}$$ Only two integer values are possible

a=2 and a=3. So corresponding "x" values are x=1 and a=3, x=2 and a=2, x=0 and a=2

where x>=0 and x<2

Applying the above process we get x=1 and a=1

where x<0 and x>=2 we get a=3 and x=-1 , a=2 and x=-2

where x<0 and x<2 we get a=1 and x=-1

Hence there are total 7 values possible