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Find the possible value or values of y in the quadratic equation 4 – 4y – y2 = 0.
A. y = –2 + 2√ 2, y = −2 − 2√ 2
B. y = –4, y = 1
C. y = –3, y = –1
D. y = 2

Respuesta :

calculista

Answer:

Option A

[tex]y=-2-2\sqrt{2}[/tex]

[tex]y=-2+2\sqrt{2}[/tex]

Step-by-step explanation:

we have

[tex]4-4y-y^{2}=0[/tex]

we know that

The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to


[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]


in this problem we have


[tex]-y^{2} -4y+4=0[/tex]  

so


[tex]a=-1\\b=-4\\c=4[/tex]


substitute

[tex]y=\frac{4(+/-)\sqrt{(-4)^{2}-4(-1)(4)}} {2(-1)}[/tex]


[tex]y=\frac{4(+/-)\sqrt{16+16}} {-2}[/tex]

[tex]y=-\frac{4(+/-)4\sqrt{2}} {2}[/tex]

[tex]y=-\frac{4+4\sqrt{2}} {2}=-2-2\sqrt{2}[/tex]

[tex]y=-\frac{4-4\sqrt{2}} {2}=-2+2\sqrt{2}[/tex]

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]