Question

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

Answer 1

Option A is correct

`y= -2 + 2√(2)` and
`y= -2 - 2√(2)`

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 = (-b \pm √(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= (-4 \pm √(4^2-4(1)(-4)))/(2(1))`

⇒
`y = (-4 \pm √(16+16))/(2)`

⇒
`y = (-4 \pm √(32))/(2)`

⇒
`y= (-4 \pm 4√(2))/(2)`

Simplify:

`y= -2 \pm 2√(2)`

Therefore, the possible values of y are:

`y= -2 + 2√(2)` and
`y= -2 - 2√(2)`

Answer 2

Option A

`y=-2-2√(2)`

`y=-2+2√(2)`

Explanation:

we have

`4-4y-y^(2)=0`

we know that

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

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

in this problem we have

`-y^(2) -4y+4=0`

so

`a=-1\\b=-4\\c=4`

substitute

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

`y=\frac{4(+/-)√(16+16)} {-2}`

`y=-\frac{4(+/-)4√(2)} {2}`

`y=-\frac{4+4√(2)} {2}=-2-2√(2)`

`y=-\frac{4-4√(2)} {2}=-2+2√(2)`