(x%5E2%2Bx%2B2)-6%3D0.jpeg "(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.