x%5E2%2B6x)%5E2-4((a-2)x%5E2%2B6x)%2B4-a%5E2%3D0.jpeg "((a-2)x^2+6x)^2-4((a-2)x^2+6x)+4-a^2=0")
Solving the Equation: ((a-2)x^2+6x)^2-4((a-2)x^2+6x)+4-a^2=0
This equation appears complex at first glance, but we can solve it using a combination of factoring, substitution, and the quadratic formula. Let's break it down step-by-step:
1. Recognizing the Pattern
Notice that the equation has a structure similar to a quadratic equation:
- (something)^2 - 4(something) + 4 - a^2 = 0
Let's substitute (a-2)x^2 + 6x with the variable 'y':
- y^2 - 4y + 4 - a^2 = 0
2. Factoring and Solving for 'y'
Now, we have a simpler quadratic equation. This can be factored into:
- (y - 2)^2 - a^2 = 0
Further factoring using the difference of squares pattern:
- [(y - 2) + a] [(y - 2) - a] = 0
This gives us two possible solutions for 'y':
- y - 2 + a = 0 => y = 2 - a
- y - 2 - a = 0 => y = 2 + a
3. Substituting Back and Solving for 'x'
Now, substitute back (a-2)x^2 + 6x for 'y' in both equations:
- (a-2)x^2 + 6x = 2 - a
- (a-2)x^2 + 6x = 2 + a
These are quadratic equations in 'x'. To solve for 'x', we can rearrange them in standard quadratic form:
- (a-2)x^2 + 6x + (a - 2) = 0
- (a-2)x^2 + 6x - (a + 2) = 0
Finally, apply the quadratic formula to find the solutions for 'x' in each equation. Remember that the solutions will depend on the value of 'a'.
4. Important Considerations
- Discriminant: The discriminant of a quadratic equation (b^2 - 4ac) determines the nature of the roots. If the discriminant is positive, there are two distinct real solutions. If it's zero, there's one repeated real solution. If it's negative, there are two complex solutions.
- Restrictions: Depending on the value of 'a', there might be restrictions on the solutions. For instance, if 'a' is equal to 2, some of the terms in the equation would become zero, leading to a simpler equation.
By following these steps, you can effectively solve the given equation and obtain the solutions for 'x' in terms of 'a'. Remember to consider the discriminant and potential restrictions on the solutions based on the value of 'a'.