Solving F(a + 2) = G(2a) Given F(x) = 5x² And G(x) = X² + 12x + 85
This article dives deep into solving a specific mathematical problem involving function composition and quadratic equations. We are given two functions, f(x) = 5x² and g(x) = x² + 12x + 85, and the challenge is to find the sum of all values of 'a' that satisfy the equation f(a + 2) = g(2a). This problem combines the concepts of function evaluation, algebraic manipulation, and solving quadratic equations. Let's break down the solution step-by-step.
1. Understanding Function Composition and the Problem
Before we jump into the calculations, let's make sure we understand the core concepts. Function composition, in simple terms, means plugging one function into another. In our case, we need to find f(a + 2) and g(2a). This means we substitute 'a + 2' into the function 'f(x)' and '2a' into the function 'g(x)'. The equation f(a + 2) = g(2a) sets up a relationship between these two composed functions, and our goal is to find the values of 'a' that make this relationship true.
The key to successfully solving this problem lies in careful substitution and simplification. We need to replace the 'x' in f(x) with '(a+2)' and the 'x' in g(x) with '(2a)'. This will give us two new expressions in terms of 'a'. Setting these expressions equal to each other will result in a quadratic equation. Solving this quadratic equation will give us the values of 'a' that satisfy the original condition. Finally, we need to calculate the sum of these values. This problem elegantly combines function evaluation with the process of solving a quadratic equation, making it a great example of how different mathematical concepts can be intertwined. The initial step, substitution, is crucial for setting up the problem correctly. Any error in substitution will propagate through the rest of the solution. Therefore, we will pay close attention to accurately replacing variables and simplifying the expressions.
2. Evaluating f(a + 2)
Let's start by evaluating f(a + 2). Remember that f(x) = 5x². To find f(a + 2), we replace 'x' with '(a + 2)':
f(a + 2) = 5(a + 2)²
Now, we need to expand the square:
(a + 2)² = (a + 2)(a + 2) = a² + 4a + 4
Substitute this back into the expression for f(a + 2):
f(a + 2) = 5(a² + 4a + 4)
Finally, distribute the 5:
f(a + 2) = 5a² + 20a + 20
This is the first part of our equation. We have successfully evaluated f(a + 2) and expressed it in terms of 'a'. The expansion of the squared term and the distribution of the constant are critical steps that require careful attention to detail. A mistake in either of these steps will lead to an incorrect expression for f(a+2), which will then affect the final solution. The process of squaring a binomial, (a+2) in this case, is a fundamental algebraic operation. Remembering the formula (x+y)² = x² + 2xy + y² can be helpful in such situations. However, if you're unsure, multiplying (a+2) by itself directly ensures accuracy. The distribution step, multiplying each term inside the parenthesis by the constant 5, is equally important. Accuracy at this stage ensures that the subsequent steps will build upon a solid foundation.
3. Evaluating g(2a)
Next, we need to evaluate g(2a). Recall that g(x) = x² + 12x + 85. To find g(2a), we replace 'x' with '2a':
g(2a) = (2a)² + 12(2a) + 85
Now, simplify each term:
(2a)² = 4a²
12(2a) = 24a
Substitute these back into the expression for g(2a):
g(2a) = 4a² + 24a + 85
We have now evaluated g(2a) and expressed it in terms of 'a'. Similar to the previous step, accuracy in substitution and simplification is paramount. Here, the substitution involves replacing 'x' with '2a'. It is crucial to remember to square the entire term '2a', not just 'a'. The term 12(2a) is a straightforward multiplication, but care must be taken to ensure it's calculated correctly. The constant term, 85, remains unchanged as it does not depend on 'x' or 'a'. This step by step breakdown allows us to isolate potential errors and ensures that we are working with the correct expression for g(2a).
4. Setting f(a + 2) Equal to g(2a) and Forming the Quadratic Equation
Now that we have expressions for both f(a + 2) and g(2a), we can set them equal to each other:
5a² + 20a + 20 = 4a² + 24a + 85
Our next goal is to rearrange this equation into the standard quadratic form, which is ax² + bx + c = 0. To do this, we subtract the right-hand side of the equation from the left-hand side:
(5a² + 20a + 20) - (4a² + 24a + 85) = 0
Distribute the negative sign and combine like terms:
5a² + 20a + 20 - 4a² - 24a - 85 = 0
(5a² - 4a²) + (20a - 24a) + (20 - 85) = 0
This simplifies to:
a² - 4a - 65 = 0
We now have a quadratic equation in the standard form. This quadratic equation is the core of the problem. Solving it will give us the values of 'a' that satisfy the initial condition f(a+2) = g(2a). The process of rearranging the equation involved careful attention to signs and combining like terms accurately. Each term needs to be correctly transferred from one side of the equation to the other. The standard form of a quadratic equation, ax² + bx + c = 0, provides a structured way to approach the problem. Once in this form, we can apply various techniques to find the solutions, such as factoring, completing the square, or using the quadratic formula. In this case, the next step involves either factoring the quadratic or applying the quadratic formula. The choice of method often depends on the specific coefficients of the equation. This step sets the stage for the final solution of the problem.
5. Solving the Quadratic Equation
We have the quadratic equation a² - 4a - 65 = 0. We can solve this equation using the quadratic formula, which is:
a = (-b ± √(b² - 4ac)) / 2a
In our equation, a = 1, b = -4, and c = -65. Substitute these values into the quadratic formula:
a = (4 ± √((-4)² - 4 * 1 * -65)) / (2 * 1)
Simplify:
a = (4 ± √(16 + 260)) / 2
a = (4 ± √276) / 2
a = (4 ± √(4 * 69)) / 2
a = (4 ± 2√69) / 2
Divide both terms in the numerator by 2:
a = 2 ± √69
So, we have two solutions for 'a':
a₁ = 2 + √69
a₂ = 2 - √69
The quadratic formula is a powerful tool for solving any quadratic equation. It guarantees a solution, even when factoring is difficult or impossible. The proper application of the formula involves careful substitution of the coefficients and accurate simplification of the resulting expression. The discriminant, b² - 4ac, under the square root, provides information about the nature of the roots. In this case, the discriminant is positive, indicating two distinct real roots. The simplification steps involve evaluating the square root and reducing the fraction. Identifying perfect square factors within the radical, such as 4 in this case, helps to simplify the expression. The final solutions are in the form of two distinct values, each involving a radical term. These values represent the 'a' coordinates where the functions f(a+2) and g(2a) intersect.
6. Finding the Sum of the Values of 'a'
The question asks for the sum of all values of 'a' that satisfy the equation. We have found two values:
a₁ = 2 + √69
a₂ = 2 - √69
To find the sum, we add these two values together:
Sum = (2 + √69) + (2 - √69)
The square root terms cancel out:
Sum = 2 + 2
Sum = 4
Therefore, the sum of all values for 'a' such that f(a + 2) = g(2a) is 4.
Alternatively, we could have used Vieta's formulas. For a quadratic equation in the form ax² + bx + c = 0, the sum of the roots is given by -b/a. In our equation, a² - 4a - 65 = 0, a = 1 and b = -4. Therefore, the sum of the roots is -(-4)/1 = 4. This provides a quicker way to find the sum of the roots without explicitly calculating them. This final step completes the solution to the problem. The key insight is that the radical terms cancel out when the two solutions are added together, leaving a simple integer value. The alternative approach using Vieta's formulas demonstrates a more efficient way to arrive at the same answer. Vieta's formulas are a valuable tool in solving quadratic equations, especially when the sum or product of the roots is required rather than the roots themselves.
7. Conclusion
In conclusion, by carefully evaluating the functions, setting up the equation, and solving the resulting quadratic, we found that the sum of all values for 'a' such that f(a + 2) = g(2a) is 4. This problem highlights the importance of understanding function composition, algebraic manipulation, and quadratic equation-solving techniques. The solution process involved a series of steps, each requiring attention to detail and accuracy. From substituting the expressions into the functions to simplifying the quadratic equation and applying the quadratic formula, each step contributed to the final answer. The alternative approach using Vieta's formulas provides a valuable shortcut for finding the sum of the roots, further demonstrating the interconnectedness of different mathematical concepts. This problem serves as a good example of how algebra and function theory can be applied to solve complex mathematical questions. The ability to break down a problem into smaller, manageable steps is crucial for success in mathematics, and this example illustrates this principle effectively.