Answer:
If you mean only one rational solution, the answer is
[tex]k_1 = 8, k_2 = -8[/tex]
If you mean at least 1 rational solution, the answer is
[tex]k\in (-\infty, -8]\cup[8, \infty)[/tex]
Step-by-step explanation:
[tex]4x^2 + kx + 4 = 0[/tex]
Let's calculate the discriminant.
[tex]\Delta = b^2 - 4ac[/tex]
[tex]\Delta = k^2 -4 \cdot 4 \cdot 4[/tex]
[tex]\Delta = k^2 -64[/tex]
Now, remember that:
[tex]\text{If } \Delta > 0 : \text{2 Real solutions}[/tex]
[tex]\text{If } \Delta = 0 : \text{1 Real solution}[/tex]
[tex]\text{If } \Delta < 0 : \text{No Real solution}[/tex]
Therefore, I will just consider the first two cases.
[tex]k^2 - 64 > 0[/tex]
and
[tex]k^2 -64 = 0[/tex]
[tex]\boxed{\text{For } k^2 - 64 > 0}[/tex]
[tex]k^2 > 64 \Longleftrightarrow k>\pm\sqrt{64} \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)[/tex]
[tex]\boxed{\text{For } k^2 - 64 = 0}[/tex]
[tex]k^2 = 64 \Longleftrightarrow k=\pm\sqrt{64} \implies k_1 = 8, k_2 = -8[/tex]
