(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2=0

3 min read Jun 17, 2024

Solving the Equation: (x^2+4x+8)^2+3x(x^2+4x+8)+2x^2=0

This equation might look intimidating at first glance, but with a clever substitution, it becomes much easier to solve. Here's how:

Let's simplify the equation by substituting a new variable. Let y = x^2 + 4x + 8.

Now our equation becomes:

y^2 + 3xy + 2x^2 = 0

This equation now looks more familiar. We can factor it as a quadratic:

(y + 2x)(y + x) = 0

This gives us two possible solutions:

Solving for y in both cases:

Now we need to substitute back our original expression for y:

We now have two quadratic equations to solve.

x^2 + 6x + 8 = 0 This factors into (x+2)(x+4) = 0, giving us solutions x = -2 and x = -4.
x^2 + 5x + 8 = 0 This equation doesn't factor easily. We can use the quadratic formula to find the solutions:

x = [-b ± √(b^2 - 4ac)] / 2a

where a = 1, b = 5, and c = 8. Plugging these values into the formula, we get:

x = [-5 ± √(5^2 - 4 * 1 * 8)] / 2 * 1 x = [-5 ± √(-7)] / 2 x = (-5 ± i√7) / 2

This gives us two complex solutions: x = (-5 + i√7) / 2 and x = (-5 - i√7) / 2

Therefore, the solutions to the original equation (x^2+4x+8)^2+3x(x^2+4x+8)+2x^2=0 are: