(x+1)(x+3)(x-4)(x-6)+24 Solution

3 min read Jun 16, 2024

Solving the Equation (x+1)(x+3)(x-4)(x-6) + 24 = 0

This problem involves a quartic equation, meaning it has a highest power of 4. While we can't always solve quartic equations directly, this particular problem has a clever solution that simplifies the equation significantly.

The Key Observation

Notice that the first part of the equation resembles the expansion of a quadratic equation. Let's manipulate it a bit:

(x+1)(x+3)(x-4)(x-6) + 24 = 0

We can rewrite this as:

[(x+1)(x-6)][(x+3)(x-4)] + 24 = 0

Now, expand each pair of brackets:

(x² - 5x - 6)(x² - x - 12) + 24 = 0

The Substitution Trick

Here's the clever part: Let's substitute y = x² - x. This substitution transforms the equation into a quadratic:

(y - 6)(y - 12) + 24 = 0

Expanding this gives us:

y² - 18y + 72 + 24 = 0

y² - 18y + 96 = 0

Solving the Quadratic

Now, we have a simple quadratic equation that we can solve using the quadratic formula or factoring:

(y - 12)(y - 8) = 0

This gives us two solutions for y:

y = 12
y = 8

Back to the Original Variable

Remember, we made the substitution y = x² - x. Now, we need to substitute back to find the solutions for x:

Case 1: y = 12

x² - x = 12

x² - x - 12 = 0

(x - 4)(x + 3) = 0

This gives us two solutions: x = 4 and x = -3

Case 2: y = 8

x² - x = 8

x² - x - 8 = 0

This equation doesn't factor easily. We can solve it using the quadratic formula:

x = [1 ± √(1² - 4(1)(-8))] / 2(1)

x = [1 ± √33] / 2

This gives us two more solutions: x = (1 + √33) / 2 and x = (1 - √33) / 2

Final Solutions

Therefore, the equation (x+1)(x+3)(x-4)(x-6) + 24 = 0 has four solutions:

x = 4
x = -3
x = (1 + √33) / 2
x = (1 - √33) / 2