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

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• 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
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