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

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.

But that answer is wrong!

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.

To find the sum of all constants k such that the equation |x^2 - 169| - x - k = 0 has three distinct real roots, we can start by analyzing the equation.

First, let's simplify the equation |x^2 - 169| - x - k = 0.

The absolute value expression, |x^2 - 169|, can be rewritten as:

1. For x^2 - 169 ≥ 0: x^2 - 169
2. For x^2 - 169 < 0: -(x^2 - 169) = -x^2 + 169

We can break down the equation |x^2 - 169| - x - k = 0 into two separate cases:

Case 1: When x^2 - 169 ≥ 0
If x^2 - 169 ≥ 0, then |x^2 - 169| = x^2 - 169.
Substituting this in the equation:

x^2 - 169 - x - k = 0

Rearranging the terms:

x^2 - x - 169 - k = 0

From this equation, we can see that the coefficients of x^2 and x are both positive, which means the quadratic equation has a concave-upward curve. For this equation to have three distinct real roots, it would mean that the discriminant (b^2 - 4ac) is greater than zero.

Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

We have a = 1, b = -1, and c = -169:

Discriminant, D = (-1)^2 - 4(1)(-169) = 1 + 676 = 677

Since the discriminant D is greater than zero, the equation has two distinct real roots for x^2 - x - 169 - k = 0.

Case 2: When x^2 - 169 < 0
If x^2 - 169 < 0, then |x^2 - 169| = -x^2 + 169.
Substituting this in the equation:

-x^2 + 169 - x - k = 0

Rearranging the terms:

-x^2 - x + 169 - k = 0

Now, we can see that the quadratic coefficients -x^2 and -x are both negative, which means the quadratic equation has a concave-downward curve. For this equation to have three distinct real roots, it would require that the discriminant (b^2 - 4ac) is also greater than zero.

Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

Again, a = -1, b = -1, and c = 169:

Discriminant, D = (-1)^2 - 4(-1)(169) = 1 + 676 = 677

So, the discriminant D is also greater than zero, and this equation has two distinct real roots.

Therefore, for both cases, we obtain two distinct real roots. To have a total of three distinct real roots for the equation |x^2 - 169| - x - k = 0, the two distinct real roots from each case must be equal to each other, resulting in a total of three distinct real roots when combined.

Let's set the two discriminants equal to each other and solve for k:

677 = 677
x^2 - x - 169 - k = -x^2 - x + 169 - k

Combining like terms:
2x^2 - 338 = 0

Dividing by 2:
x^2 - 169 = 0

Factoring:
(x - 13)(x + 13) = 0

So, we have two solutions for x: x = 13 and x = -13.

Substituting these x-values into either case equation and solving for k:

For x = 13:
x^2 - x - 169 - k = 0
13^2 - 13 - 169 - k = 0
169 - 13 - 169 - k = 0
-13 - k = 0
k = -13

For x = -13:
x^2 - x - 169 - k = 0
(-13)^2 - (-13) - 169 - k = 0
169 + 13 - 169 - k = 0
13 - k = 0
k = 13

Hence, we have two values for k: k = -13 and k = 13.

The sum of all constants k that satisfy the given condition is:

-13 + 13 = 0

Therefore, the sum of all constants k is zero.