Question

For what values of a does the equation ax2+x+4=0 have only one real solution?

Answer 1

a ≤ 1/16, or (-∞, 1/16]

Explanation:

ax² + x + 4 = 0

To have real solutions discriminant of the equation should be D ≥ 0.

ax² + bx + c = 0 , D = b² - 4ac

ax² + x + 4 = 0, a, b = 1, c = 4, so D = 1² - 4*a*4 = 1 - 16a

D ≥ 0

1 - 16a ≥ 0

16a ≤ 1

a ≤ 1/16, or (-∞, 1/16]

Answer 2

1/16

Explanation:

To have one real solution, the discriminant must be 0.

b² − 4ac = 0

1² − 4a(4) = 0

1 − 16a = 0

a = 1/16