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

1/16

Step-by-step explanation:

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

b² − 4ac = 0

1² − 4a(4) = 0

1 − 16a = 0

a = 1/16

a ≤ 1/16, or (-∞, 1/16]

Step-by-step 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: i think it's a

step-by-step explanation:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11