(x^2+x)^2+4(x^2+x)-12=0

3 min read Jun 17, 2024

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.