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

2 min read Jun 17, 2024

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

This equation might look intimidating at first, but we can solve it by using a clever substitution and applying our knowledge of quadratic equations.

Let's simplify the equation by substituting a new variable. Let:

y = x^2 + 3x

Now our equation becomes:

y^2 - 2y - 8 = 0

This is a much more familiar quadratic equation!

We can solve this quadratic equation using the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / 2a

Where a = 1, b = -2, and c = -8.

Substituting these 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:

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

We have two quadratic equations to solve:

1. x^2 + 3x - 4 = 0

This factors nicely:

(x + 4)(x - 1) = 0

Therefore, x = -4 or x = 1.

2. x^2 + 3x + 2 = 0

This also factors nicely:

(x + 1)(x + 2) = 0

Therefore, x = -1 or x = -2.

The solutions to the equation (x^2 + 3x)^2 - 2(x^2 + 3x) - 8 = 0 are: