X(x+1)(x+2)(x+3)=24

To solve the equation \( x(x+1)(x+2)(x+3) = 24 \), we can start by letting \( y = x(x+1)(x+2)(x+3) \). This equation is a polynomial expression of degree 4, meaning it could have up to 4 real roots. However, to make it easier to handle, we can start by simplifying the left side. Notice that: \[ x(x+1)(x+2)(x+3) = (x^2 + 3x)(x^2 + 3x + 2) \] Let \( z = x^2 + 3x \). Then the equation becomes: \[ z(z + 2) = 24 \] Expanding this gives: \[ z^2 + 2z - 24 = 0 \] Now, we can use the quadratic formula to solve for \( z \): \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2} = \frac{-2 \pm 10}{2} \] Calculating the two possible values for \( z \): 1. \( z = \frac{8}{2} = 4 \) 2. \( z = \frac{-12}{2} = -6 \) Now, we revert back to \( x \) using \( z = x^2 + 3x \): 1. For \( z = 4 \): \[ x^2 + 3x - 4 = 0 \] Using the quadratic formula: \[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \] This gives us: - \( x = \frac{2}{2} = 1 \) - \( x = \frac{-8}{2} = -4 \) 2. For \( z = -6 \): \[ x^2 + 3x + 6 = 0 \] Using the quadratic formula again: \[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(6)}}{2(1)} = \frac{-3 \pm \sqrt{9 - 24}}{2} = \frac{-3 \pm \sqrt{-15}}{2} \] Since this gives a negative discriminant, we have no real solutions from this equation. Thus, the real solutions to the original equation \( x(x+1)(x+2)(x+3) = 24 \) are: \[ \boxed{1} \quad \text{and} \quad \boxed{-4} \]