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8 October, 20:05

Write the quadratic equation whose coefficient with x^2 to 1 and roots are √3-1 divided by 2 and √3+1 divided by 2

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Answers (2)
  1. 8 October, 21:25
    0
    1x^2-sqrt (3) b+1/2=0 or

    2x^2 - 2sqrt (3) b+1=0

    Step-by-step explanation:

    ax^2 + bx+c=0

    a=1, b, c=?

    x1 = (sqrt (3) - 1) / 2

    x2 = (sqrt (3) + 1) / 2

    x1+x2=-b/a

    x1 * x2 = c/a

    x1 + x2=

    (sqrt (3) - 1) / 2 + (sqrt (3) + 1) / 2=

    2sqrt (3) / 2=

    sqrt (3) = -b/1

    b=-sqrt (3)

    x1*x2 =

    ((sqrt (3) - 1) / 2) * ((sqrt (3) + 1) / 2) =

    ((sqrt (3)) ^2 - 1^2) / 4=

    (3-1) / 4=

    2/4=

    1/2=c/1

    c=1/2

    ax^2 + bx+c=0

    1x^2-sqrt (3) b+1/2=0 / *2

    2x^2 - 2sqrt (3) b+1=0
  2. 8 October, 22:08
    0
    Now that we have both of these zero terms, we can multiply them to get a standard form.

    f (x) = (x - 6) (x + 4)

    And while this will give us the zeros we need, it will no give us the lead coefficient. So we must multiply by the desired lead coefficient.

    f (x) = 5 (x - 6) (x + 4)

    f (x) = 5 (x^2 - 6x + 4x - 24)

    f (x) = 5 (x^2 - 2x - 24)

    f (x) = 5x^2 - 10x - 120
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