Question

For which value of k will x² -3x -k can be factored completely

Answer 1

To determine the value of \( k \) for which the quadratic \( x^2 - 3x - k \) can be factored completely, we will use the fact that a quadratic of the form \( ax^2 + bx + c \) can be factored as \( (x - p)(x - q) \) where \( p \) and \( q \) are the roots of the quadratic.

The quadratic can be factored completely when the discriminant \( D \) is a perfect square. The discriminant for a quadratic \( ax^2 + bx + c \) is given by:

\[
D = b^2 - 4ac
\]

In our case, we have \( a = 1 \), \( b = -3 \), and \( c = -k \). Thus, the discriminant becomes:

\[
D = (-3)^2 - 4(1)(-k) = 9 + 4k
\]

We want \( 9 + 4k \) to be a perfect square. Let’s set \( 9 + 4k = m^2 \) for some integer \( m \):

\[
4k = m^2 - 9
\]
\[
k = \frac{m^2 - 9}{4}
\]

To ensure \( k \) is an integer, \( m^2 - 9 \) must be divisible by 4. Since \( 9 \equiv 1 \pmod{4} \), \( m^2 \) must also be congruent to \( 1 \pmod{4} \), which means \( m \) must be odd. The odd integers can be represented as \( m = 2n + 1 \) for some integer \( n \).

Let’s substitute \( m = 2n + 1 \):

\[
k = \frac{(2n + 1)^2 - 9}{4}
\]
\[
= \frac{4n^2 + 4n + 1 - 9}{4}
\]
\[
= \frac{4n^2 + 4n - 8}{4}
\]
\[
= n^2 + n - 2
\]

The values for \( n \) lead to integer values for \( k \):

- If \( n = 0 \): \( k = 0^2 + 0 - 2 = -2 \)
- If \( n = 1 \): \( k = 1^2 + 1 - 2 = 0 \)
- If \( n = 2 \): \( k = 2^2 + 2 - 2 = 4 \)
- If \( n = -1 \): \( k = (-1)^2 - 1 - 2 = -2 \) (same as above)
- If \( n = -2 \): \( k = (-2)^2 - 2 - 2 = 0 \) (same as above)

Thus, the integer values for \( k \) that allow the quadratic to be factored completely are \( k = -2 \) and \( k = 0 \).

- For \( k = -2 \):
\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]

- For \( k = 0 \):
\[
x^2 - 3x = x(x - 3)
\]

Therefore, the values of \( k \) for which \( x^2 - 3x - k \) can be factored completely are:

\[
\boxed{-2} \text{ and } \boxed{0}
\]