(x^2-4x)^2+7(x^2-4x)+12=0

2 min read Jun 17, 2024
(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

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