Question

Find the value(s) of k such that 4x² + kx + 4 = 0 has 1 rational solution

Answer 1

Answer:

If you mean only one rational solution, the answer is

$k_1 = 8, k_2 = -8$

If you mean at least 1 rational solution, the answer is

$k\in (-\infty, -8]\cup[8, \infty)$

Step-by-step explanation:

$4x^2 + kx + 4 = 0$

Let's calculate the discriminant.

$\Delta = b^2 - 4ac$

$\Delta = k^2 -4 \cdot 4 \cdot 4$

$\Delta = k^2 -64$

Now, remember that:

$\text{If } \Delta > 0 : \text{2 Real solutions}$

$\text{If } \Delta = 0 : \text{1 Real solution}$

$\text{If } \Delta < 0 : \text{No Real solution}$

Therefore, I will just consider the first two cases.

$k^2 - 64 > 0$

and

$k^2 -64 = 0$

$\boxed{\text{For } k^2 - 64 > 0}$

$k^2 > 64 \Longleftrightarrow k>\pm\sqrt{64} \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)$

$\boxed{\text{For } k^2 - 64 = 0}$

$k^2 = 64 \Longleftrightarrow k=\pm\sqrt{64} \implies k_1 = 8, k_2 = -8$