Question

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

Answer 1

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)`

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√(64) \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)`

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

`k^2 = 64 \Longleftrightarrow k=\pm√(64) \implies k_1 = 8, k_2 = -8`