Solving the Quadratic Equation: (x^2 + x)^2 + 4(x^2 + x)  12 = 0
This equation might look intimidating at first glance, but we can solve it by employing a clever substitution and our knowledge of quadratic equations. Let's break down the steps:
1. Substitution
Notice that the expression (x^2 + x)
appears repeatedly. Let's make our lives easier by substituting it with a new variable, say y
:
 Let y = x^2 + x
Now our equation transforms into:
 y^2 + 4y  12 = 0
2. Solving the Quadratic Equation
This is a standard quadratic equation in the form ay^2 + by + c = 0
. We can solve it using the quadratic formula:
 y = (b ± √(b^2  4ac)) / 2a
In our case, a = 1, b = 4, and c = 12. Plugging these values into the quadratic formula:
 y = (4 ± √(4^2  4 * 1 * 12)) / (2 * 1)
 y = (4 ± √(64)) / 2
 y = (4 ± 8) / 2
This gives us two possible solutions for y
:
 y1 = 2
 y2 = 6
3. Back Substitution
Now we need to substitute back x^2 + x
for y
and solve for x
:

For y1 = 2:
 x^2 + x = 2
 x^2 + x  2 = 0
 (x + 2)(x  1) = 0
 x = 2 or x = 1

For y2 = 6:
 x^2 + x = 6
 x^2 + x + 6 = 0
This equation doesn't factor easily. We can use the quadratic formula again to find the solutions:
 x = (1 ± √(1^2  4 * 1 * 6)) / (2 * 1)
 x = (1 ± √(23)) / 2
 x = (1 ± i√23) / 2 (where 'i' is the imaginary unit, √1)
4. Solutions
Therefore, the solutions to the equation (x^2 + x)^2 + 4(x^2 + x)  12 = 0 are:
 x = 2
 x = 1
 x = (1 + i√23) / 2
 x = (1  i√23) / 2
We have two real solutions and two complex solutions.