(x^2+x+1)(x^2+x+2)-6=0

3 min read Jun 17, 2024
(x^2+x+1)(x^2+x+2)-6=0

Solving the Equation (x^2 + x + 1)(x^2 + x + 2) - 6 = 0

This equation might look intimidating at first, but it can be solved using a clever substitution and factoring. Here's how:

1. Substitution:

Let's simplify the equation by making a substitution. Let y = x^2 + x. Now our equation becomes:

(y + 1)(y + 2) - 6 = 0

2. Expanding and Simplifying:

Expand the left side of the equation:

y^2 + 3y + 2 - 6 = 0

Simplify:

y^2 + 3y - 4 = 0

3. Factoring:

Now we have a simple quadratic equation. Factor it:

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

4. Solving for y:

This gives us two possible solutions for y:

  • y + 4 = 0 => y = -4
  • y - 1 = 0 => y = 1

5. Substituting Back:

Now we need to substitute back x^2 + x for y in both cases:

  • Case 1: x^2 + x = -4
  • Case 2: x^2 + x = 1

6. Solving for x:

Case 1: x^2 + x + 4 = 0

This quadratic equation doesn't factor easily. We can use the quadratic formula to find the solutions:

x = [-b ± √(b^2 - 4ac)] / 2a

Where a = 1, b = 1, and c = 4.

After plugging in the values, we find that the solutions for this case are complex numbers.

Case 2: x^2 + x - 1 = 0

Again, we can use the quadratic formula:

x = [-b ± √(b^2 - 4ac)] / 2a

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

Solving this gives us two real solutions for x.

7. The Solutions:

Therefore, the solutions to the equation (x^2 + x + 1)(x^2 + x + 2) - 6 = 0 are:

  • Two real solutions found by solving x^2 + x - 1 = 0 using the quadratic formula.
  • Two complex solutions found by solving x^2 + x + 4 = 0 using the quadratic formula.

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