21 Suppose the equation ||$x-a|-b|=2008$ has 3 distinct real roots and $a \neq 0$. Find the value of $b$.
21. Answer: 2008
The equation is equivalent to $|x-a|=b \pm 2008$.
Case 1: If $b2008$, then both $|x-a|=b-2008$ and $|x-a|=b+2008$ has 2 real roots, which gives 4 distinct real roots $a \pm(b-2008)$ and $a \pm(b+2008)$, since $a+b+2008>a+b-2008>a-b+2008>a-b-2008$.
Case 3: If $b=2008$, then $|x-a|=b-2008=0$ has only 1 real root $x=a$, and $|x-a|=b+2008=4016$ has 2 real roots $x=a \pm 4016$.
Hence $b=2008$.