Answer:
Option A is correct
\(y= -2 + 2\sqrt{2}\) and \(y= -2 - 2\sqrt{2}\)
Step-by-step explanation:
A quadratic equation is in the form of:
\(ax^2+bx+c = 0\),......[1] then
the solution of the equation is given by:
\(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) .....[2]
As per the statement:
Given the equation:
\(4 -4y-y^2 = 0.\)
We can write this as:
\(y^2+4y-4 = 0\)
On comparing with [1] we have;
a = 1, b = 4 and c = -4
Substitute these in [2] we have;
\(y= \frac{-4 \pm \sqrt{4^2-4(1)(-4)}}{2(1)}\)
⇒ \(y = \frac{-4 \pm \sqrt{16+16}}{2}\)
⇒ \(y = \frac{-4 \pm \sqrt{32}}{2}\)
⇒ \(y= \frac{-4 \pm 4\sqrt{2}}{2}\)
Simplify:
\(y= -2 \pm 2\sqrt{2}\)
Therefore, the possible values of y are:
\(y= -2 + 2\sqrt{2}\) and \(y= -2 - 2\sqrt{2}\)
