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

Jul 13, 2018

$y = - 2 \pm \sqrt{2} , \text{ } \frac{1}{2} \pm \frac{\sqrt{7} i}{2}$

#### Explanation:

Given: ${\left(y + \frac{2}{y}\right)}^{2} + 3 y + \frac{6}{y} = 4$

This is one way to solve. Use ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

${y}^{2} + 2 \cancel{y} \left(\frac{2}{\cancel{y}}\right) + \frac{4}{y} _ 2 + 3 y + \frac{6}{y} = 4$

${y}^{2} + 4 + \frac{4}{y} _ 2 + 3 y + \frac{6}{y} = 4$

Multiply both sides by ${y}^{2}$ to eliminate the fractions:

${y}^{4} + 4 {y}^{2} + 4 + 3 {y}^{3} + 6 y = 4 {y}^{2}$

Add like terms and put in descending order:

${y}^{4} + 3 {y}^{3} + 6 y + 4 = 0$

Factor:

Can't use group factoring.

Use $\left({y}^{2} + a y + b\right) \left({y}^{2} + c y + d\right) = {y}^{4} + 3 {y}^{3} + 6 y + 4$

${y}^{4} + \left(a + c\right) {y}^{3} + \left(d + a c + b\right) {y}^{2} + \left(a d + b c\right) y + b d = {y}^{4} + 3 {y}^{3} + 6 y + 4$

Solve the system:

$a + c = 3 \text{ }$ the coefficient of the ${y}^{3}$ term
$d + a c + b = 0 \text{ }$ because there is no ${y}^{2}$ term
$a d + b c = 6 \text{ }$ the coefficient of the $y$ term
$b d = 4$

Start with the possibilities for $b d = \left(2 , 2\right) , \left(4 , 1\right) , \left(1 , 4\right)$

If $b = 2 , d = 2$, then from the 2nd equation: $a c = - 4$

Try $a = - 1 , c = 4 \text{ }$ works for all equations!

Factored: $\text{ } \left({y}^{2} - y + 2\right) \left({y}^{2} + 4 y + 2\right) = 0$

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)

Jul 13, 2018

${y}_{1} = \frac{1 + i \sqrt{7}}{2}$, ${y}_{2} = \frac{1 - i \sqrt{7}}{2}$, ${y}_{3} = - 2 + \sqrt{2}$ and ${y}_{4} = - 2 - \sqrt{2}$

#### Explanation:

${\left(y + \frac{2}{y}\right)}^{2} + 3 y + \frac{6}{y} = 4$

${\left(y + \frac{2}{y}\right)}^{2} + 3 \cdot \left(y + \frac{2}{y}\right) = 4$

After setting $x = y + \frac{2}{y}$, this equation became

${x}^{2} + 3 x = 4$

${x}^{2} + 3 x - 4 = 0$

$\left(x + 4\right) \cdot \left(x - 1\right) = 0$, so ${x}_{1} = 1$ and ${x}_{2} = - 4$

a) For $x = 1$,

$y + \frac{2}{y} = 1$

${y}^{2} + 2 = y$

${y}^{2} - y + 2 = 0$, consequently ${y}_{1} = \frac{1 + i \sqrt{7}}{2}$ and ${y}_{2} = \frac{1 - i \sqrt{7}}{2}$

b) For $x = - 4$,

$y + \frac{2}{y} = - 4$

${y}^{2} + 2 = - 4 y$

${y}^{2} + 4 y + 2 = 0$, consequently ${y}_{3} = - 2 + \sqrt{2}$ and ${y}_{4} = - 2 - \sqrt{2}$