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

Question

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

in progress 0
Josie 2 weeks 2021-10-11T18:35:24+00:00 2 Answers 0

Answers ( )

    0
    2021-10-11T18:36:38+00:00

    Answer:

    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

    0
    2021-10-11T18:37:21+00:00

    Answer:

    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]

Leave an answer

Browse
Browse

27:3+15-4x7+3-1=? ( )