Show that the equation (k + 1)x2 − 2x − k = 0 has a solution for all values of k

To show that the equation (k + 1)x^2 - 2x - k = 0 has a solution for all values of k, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = (k + 1), b = -2, and c = -k. Substituting these values into the quadratic formula, we get:

x = (-(-2) ± √((-2)^2 - 4(k + 1)(-k))) / (2(k + 1))

Simplifying further, we have:

x = (2 ± √(4 + 4k^2 + 4k)) / (2(k + 1)) x = (2 ± 2√(1 + k^2 + k)) / (2(k + 1)) x = (1 ± √(1 + k^2 + k)) / (k + 1)

Now, we need to show that this expression for x is valid for all values of k. To do this, we can consider the denominator (k + 1). For any value of k, the denominator will never be zero, as adding 1 to any number will always result in a non-zero value. Therefore, the expression for x is defined for all values of k.

In conclusion, the equation (k + 1)x^2 - 2x - k = 0 has a solution for all values of k, given by x = (1 ± √(1 + k^2 + k)) / (k + 1).