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