What is the sum of all the constants k such that |x^2−169|−x−k=0 has three distinct real roots?

Answer 1

if |x| >= 13, |x^2-169| = x^2-169
In that case, (x-13)(x+13) - (x+k) = 0

If k = 13,

(x-13)(x+13) - (x+13) = 0
(x-12)(x+13) = 0

Also, if x < 13,
-(x^2-169) - (x+13) = 0
(x-14)(x-13) = 0

So, if k=13, x=-13,12,14

For other values of k, we will still have two roots near x=13, but we will lose the single root at x = -13, having either none or two there.

So, the answer is 13.

Answer 2

But that answer is wrong!

Answer 3

Hmmm. Can't think of much else right now. Maybe you can see the flaw in my logic. May have to get back to you on that.