find the possible value of y in the quadratic equation 4-4y-y^2=0?

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7 Answers

  • y^2 + 4y = 4

    y^2 + 4y + 4 = 8

    (y + 2)^2 = 8

    y + 2 = +/- 2 sqrt(2)

    y = -2 +/- 2 sqrt(2)

  • Assuming y2 is y^2 (y-squared), the equation isn’t: merely improve the words contained in the brackets and also you get: y^2 + 4y – y^2 = 6 Which simplifies to: 4y = 6 which isn’t a quadratic equation. yet when y2 isn’t y-squared yet truly 2y (i.e. 2 more effective with the help of y) then the equation is quadratic: y^2 + 2y = 6. desire this facilitates. suited answer? 😛

  • 4 – 4y – y^2 = 0

    -y^2 – 4y + 4 = 0

    -(y^2 + 4y – 4) = 0

    y^2 – 4y – 4 = 0

    y = (-b ± sqrt(b^2 – 4ac))/(2a)

    y = (-(-4) ± sqrt((-4)^2 – 4(1)(-4)))/(2(1))

    y = (4 ± sqrt(16 + 16))/2

    y = (4 ± sqrt(32))/2

    y = (4 ± sqrt(16 * 2))/2

    y = (4 ± 4sqrt(2))/2

    ANS : 2 + 2sqrt(2) or 2 – 2sqrt(2)

  • -(y^2+4y-4)=0 values are -4+4(2)^1/2/2 and -4-4(2)^1/2//

    =-2+2rt2 and -2-2rt2

  • factorize it by using the quadratic formula:

    (-b +- sqrt(D))/2a, D is the determinant formula I gave you on the other question.

    or using the completing the squares method

    Source(s): do your own homework.
  • corey, what part of your math assignment are you doing? (Just posting problems is not doing your assignment). I thought you knew that…

  • 0.828 or-4.828 (correct upto 3 digit)

    Source(s): x=-b+/-sqrt(b^2-4ac)/21

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