What are the zeros of the quadratic function f(x) = 6×2 – 24x + 1? © x=-2+, orx=-2- • x=2 +, or x = 2-F © x=-2+, or x=-2-F

Question

What are the zeros of the quadratic function f(x) = 6×2 – 24x + 1?
© x=-2+, orx=-2-
• x=2 +, or x = 2-F
© x=-2+, or x=-2-F
• x=2+2 orx=2-7

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4 weeks 2021-09-23T21:13:52+00:00 2 Answers 0

Answers ( )

  1. Charlotte
    0
    2021-09-23T21:14:52+00:00

    Answer:

    x1 = 3.95

    x2= 0.0425

    Step-by-step explanation:

    Hi, to answer this question we have to apply the quadratic formula:

    For: ax2+ bx + c

    x =[ -b ± √b²-4ac] /2a

    Replacing with the values given:

    x =[ -(-24) ± √(-24)²-4(6)1] /2(6)

    x = [ 24 ± √576 -24] /12

    x = [ 24 ± √552] /12

    x = [ 24 ± 23.49] /12

    Positive:

    x = [ 24 + 23.49] /12 = 47.49 /12 = 3.95

    Negative:

    x = [ 24 – 23.49] /12 = 0.51 /12 = 0.0425

    Feel free to ask for more if needed or if you did not understand something.

  2. Charlotte
    0
    2021-09-23T21:15:18+00:00

    Answer:

    x_1 \approx 3.958

    x_{2} \approx 0.042

    f(x) = (x - 3.958)\cdot (x-0.042)

    Step-by-step explanation:

    The zeros of the quadratic function are obtained after some algebraic manipulation:

    f(x) = 6\cdot x^{2} - 24\cdot x + 1

    The roots of the second order polynomial are, respectively:

    x_{1,2} = \frac{24 \pm \sqrt{576-4\cdot (6)\cdot (1)}}{2\cdot (6)}

    x_{1,2} \approx 2 \pm 1.958

    x_1 \approx 3.958

    x_{2} \approx 0.042

    f(x) = (x - 3.958)\cdot (x-0.042)

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