%5E2%2B7(x%5E2-4x)%2B12%3D0.jpeg "(x^2-4x)^2+7(x^2-4x)+12=0")
Solving the Quadratic Equation: (x^2-4x)^2+7(x^2-4x)+12=0
This equation might look intimidating at first glance, but it can be solved effectively by using a simple substitution technique.
Step 1: Substitution
Let's make the substitution y = x² - 4x. This allows us to rewrite the equation as:
y² + 7y + 12 = 0
This is now a standard quadratic equation in terms of 'y'.
Step 2: Factoring
We can factor the quadratic equation:
(y + 3)(y + 4) = 0
This gives us two possible solutions for 'y':
- y = -3
- y = -4
Step 3: Back-Substitution
Now, we need to substitute back the original expression for 'y' to get equations in terms of 'x':
- x² - 4x = -3
- x² - 4x = -4
Step 4: Solving for x
Let's solve each equation for 'x':
For x² - 4x = -3:
- x² - 4x + 3 = 0
- (x - 1)(x - 3) = 0
- x = 1 or x = 3
For x² - 4x = -4:
- x² - 4x + 4 = 0
- (x - 2)(x - 2) = 0
- x = 2 (double root)
Conclusion
Therefore, the solutions to the equation (x²-4x)²+7(x²-4x)+12=0 are:
- x = 1
- x = 2
- x = 3