Question

Solve `(y + 2/y)^2 + 3y + 6/y = 4`?

Answer 1

`y = -2 +-sqrt(2), " "1/2 +- (sqrt(7)i)/2`

Explanation:

Given: `(y + 2/y)^2 + 3y +6/y = 4`

This is one way to solve. Use `(a+b)^2 = a^2+2ab+ b^2`

`y^2 + 2cancel(y)(2/cancel(y))+4/y_2 + 3y +6/y = 4`

`y^2 + 4+4/y_2 + 3y +6/y = 4`

Multiply both sides by `y^2` to eliminate the fractions:

`y^4 + 4y^2 + 4 + 3y^3 +6y = 4y^2`

Add like terms and put in descending order:

`y^4 + 3y^3 + 6y + 4 = 0`

Factor:

Can't use group factoring.

Use `(y^2 +ay + b)(y^2 + cy + d) = y^4 + 3y^3 + 6y + 4`

`y^4 + (a+c)y^3 + (d+ac+b)y^2 + (ad + bc)y + bd = y^4 + 3y^3 + 6y + 4`

Solve the system:

`a + c = 3" "` the coefficient of the `y^3` term
`d + ac + b = 0" "` because there is no `y^2` term
`ad + bc = 6" "` the coefficient of the `y` term
`bd = 4`

Start with the possibilities for `bd = (2, 2), (4, 1), (1, 4)`

If `b = 2, d = 2`, then from the 2nd equation: `ac = -4`

Try `a = -1, c = 4" "` works for all equations!

Factored: `" "(y^2 - y + 2)(y^2 + 4y + 2) = 0`

Solve each trinomial either by completing the square or using the quadratic formula:

`y^2 - y + 2 = 0; " "y^2 + 4y + 2 = 0`

`y = (1+- sqrt(1-4(1)(2)))/2; " " y = (-4+- sqrt(16-4(1)(2)))/2`

`y = (1+- sqrt(7)i)/2; " " y = -2 +-sqrt(8)/2 = -2 +- sqrt(2)`

Answer 2

`y_1=(1+isqrt7)/2`, `y_2=(1-isqrt7)/2`, `y_3=-2+sqrt2` and `y_4=-2-sqrt2`

Explanation:

`(y+2/y)^2+3y+6/y=4`

`(y+2/y)^2+3*(y+2/y)=4`

After setting `x=y+2/y`, this equation became

`x^2+3x=4`

`x^2+3x-4=0`

`(x+4)*(x-1)=0`, so `x_1=1` and `x_2=-4`

`a)` For `x=1`,

`y+2/y=1`

`y^2+2=y`

`y^2-y+2=0`, consequently `y_1=(1+isqrt7)/2` and `y_2=(1-isqrt7)/2`

`b)` For `x=-4`,

`y+2/y=-4`

`y^2+2=-4y`

`y^2+4y+2=0`, consequently `y_3=-2+sqrt2` and `y_4=-2-sqrt2`