jmc

Question

Let k be a real number, such that both roots of x^2 - 2kx + k^2 + k - 5 = 0are real, and they are less than 5. Find all possible values of k.

Accepted Answer

Since both roots are real, the discriminant must be nonnegative:

(- 2 k)^{2} - 4 (k^{2} + k - 5) \geq 0.

This simplifies to

20 - 4 k \geq 0,

so

k \leq 5.

Let

y = x^{2} - 2 k x + k^{2} + k - 5 = (x - k)^{2} + k - 5.

Thus, parabola opens upward, and its vertex is

(k, k - 5) .

If

k = 5,

then the quadratic has a double root of

x = 5,

so we must have

k < 5.

Then the vertex lies to the left of the line

x = 5.

Also, for both roots to be less than 5, the value of the parabola at

x = 5

must be positive. Thus,

25 - 10 k + k^{2} + k - 5 > 0.

Then

k^{2} - 9 k + 20 > 0,

or

(k - 4) (k - 5) > 0.

Since

k < 5,

we must have

k < 4.

Thus, both roots are less than 5 when

k \in (- \infty, 4) .