Find the possible value or values of y in the quadratic equation 4 - 4y - y² = 0.

Solution:

Quadratic equation 4 - 4y - y² = 0 [Given]

It can be written as

- y² - 4y + 4 = 0

Using the quadratic formula

\( y=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \)From the equation we know that

a = - 1, b = - 4 and c = 4

\( y=\frac{-(-4)\pm \sqrt{(-4)^{2}-4\times -1\times 4}}{2\times -1} \)

By further calculation

\( y=\frac{4\pm \sqrt{16+16}}{-2} \)

\( y=\frac{4\pm 4\sqrt{2}}{-2} \)

Taking out 2 as common

\( y=\frac{2(2\pm 2\sqrt{2})}{-2} \)

So we get

y = - (2 ± 2√2)

y = - (2 + 2√2) and y = - (2 - 2√2)

y = - 2 - 2√2 and y = - 2 + 2√2

Therefore, the possible values of y are - 2 - 2√2 and - 2 + 2√2.

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Find the possible value or values of y in the quadratic equation 4 - 4y - y² = 0.

Summary:

The possible value or values of y in the quadratic equation 4 - 4y - y² = 0 are - 2 - 2√2 and - 2 + 2√2.