%5E4%2B2(x%2B3)%5E2-8%3D0.jpeg "(x+3)^4+2(x+3)^2-8=0")
Solving the Equation (x+3)^4 + 2(x+3)^2 - 8 = 0
This equation might look intimidating at first, but we can solve it using a clever substitution and the quadratic formula. Here's how:
1. Substitution
Let's simplify the equation by substituting a new variable. Let y = (x + 3)^2. This allows us to rewrite the equation as:
y^2 + 2y - 8 = 0
2. Quadratic Formula
Now we have a simple quadratic equation. We can solve for y using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
Where a = 1, b = 2, and c = -8. Plugging in the values, we get:
y = (-2 ± √(2^2 - 4 * 1 * -8)) / 2 * 1
y = (-2 ± √(36)) / 2
y = (-2 ± 6) / 2
This gives us two possible solutions for y:
y1 = 2
y2 = -4
3. Solving for x
Now we need to substitute back to find the values of x. Remember that y = (x + 3)^2.
For y1 = 2:
(x + 3)^2 = 2
- Take the square root of both sides: x + 3 = ±√2
- Isolate x: x = -3 ± √2
For y2 = -4:
(x + 3)^2 = -4
- This equation has no real solutions because the square of a real number cannot be negative.
4. Solution
Therefore, the solutions to the original equation (x+3)^4 + 2(x+3)^2 - 8 = 0 are:
x = -3 + √2
x = -3 - √2