Solving Cubic Equations Arithmetic Progression Problem Step-by-Step

Hey everyone! Let's dive into a cool math problem from Colegio Ciencias y Humanidades 5. This one involves finding the value of 'm' in a cubic equation, given that its roots are in arithmetic progression. Sounds interesting, right? We'll break it down step by step so you can follow along easily. Let's get started!

Understanding the Problem: The Core of the Equation

In this challenging math problem, we're given the cubic equation -x³ + 15x² + mx - 45 = 0. The key piece of information here is that the roots of this equation form an arithmetic progression. Remember, an arithmetic progression is a sequence where the difference between consecutive terms is constant. For example, 2, 4, 6 is an arithmetic progression with a common difference of 2. This crucial detail is our starting point for solving the problem. Our mission, should we choose to accept it (and we do!), is to find the value of 'm'. To achieve this, we'll need to leverage the properties of arithmetic progressions and the relationships between the roots and coefficients of a polynomial equation. Think of the roots as the hidden keys to unlocking the value of 'm'. We need to find a way to connect these roots, which are in arithmetic progression, to the coefficients of the equation, particularly 'm'. This is where our algebraic skills will come into play. We'll be using Vieta's formulas, which provide a direct link between the roots and coefficients of a polynomial. By applying these formulas, we can create equations that involve the roots and 'm', allowing us to solve for the unknown. It's like detective work, where each piece of information, each property, and each formula brings us closer to the solution. We'll be carefully piecing together these elements to reveal the numerical value of 'm', the final piece of the puzzle. So, gear up your thinking caps, and let's embark on this mathematical adventure. We're about to see how the seemingly simple fact that the roots are in arithmetic progression can lead us to a specific numerical answer, demonstrating the elegant connections within the world of algebra.

Setting Up the Roots: Arithmetic Progression in Action

So, we know the roots are in arithmetic progression. Let's represent these roots in a way that reflects this. A clever way to do this is to let the roots be a - d, a, and a + d, where 'a' is the middle term and 'd' is the common difference. This representation is super helpful because it neatly captures the arithmetic progression property. Imagine these roots sitting on a number line, equally spaced apart. The beauty of this setup is that when we sum these roots, the 'd' terms will cancel out, leaving us with a simpler expression involving only 'a'. This is a strategic move that will simplify our calculations later on. Now, why is this important? Well, the relationships between the roots and coefficients of a polynomial equation, given by Vieta's formulas, involve sums and products of the roots. By representing the roots in this way, we're setting ourselves up to use Vieta's formulas effectively. We're essentially creating a mathematical lever that will allow us to lift the value of 'm' out of the equation. Think of it like this: we've transformed the problem from dealing with three separate, potentially complicated roots to dealing with just two variables, 'a' and 'd'. This is a significant simplification. But remember, our ultimate goal is to find 'm'. So, how does setting up the roots in this way help us get there? The answer lies in the connection between these roots and the coefficients of the cubic equation. By using Vieta's formulas, we can relate the sum and product of these roots to the coefficients, including 'm'. This will give us equations that we can solve to find 'a', 'd', and ultimately, 'm'. It's like building a bridge between the roots and the coefficients, with Vieta's formulas acting as the bridge's supports. So, with our roots neatly set up in arithmetic progression form, we're ready to take the next step and apply Vieta's formulas. This is where the real algebraic magic begins to happen, so let's keep our focus sharp and prepare to dive into the equations that will lead us to the solution.

Vieta's Formulas: Unveiling the Root-Coefficient Connection

Vieta's formulas are the secret sauce here. They tell us how the coefficients of a polynomial equation relate to the sums and products of its roots. For a cubic equation like ours, -x³ + 15x² + mx - 45 = 0, Vieta's formulas give us the following relationships:

• Sum of roots: (a - d) + a + (a + d) = - (coefficient of x²) / (coefficient of x³) = -15 / -1 = 15
• Sum of pairwise products of roots: (a - d)a + (a - d)(a + d) + a(a + d) = (coefficient of x) / (coefficient of x³) = m / -1 = -m
• Product of roots: (a - d) * a * (a + d) = - (constant term) / (coefficient of x³) = - (-45) / -1 = -45

These formulas are like a set of mathematical lenses that allow us to see the hidden connections within the equation. Each formula provides a different perspective, revealing a different aspect of the relationship between the roots and coefficients. The sum of the roots formula, for instance, gives us a direct link between the roots and the coefficient of the x² term. The sum of pairwise products formula connects the roots to the coefficient of the x term, which is 'm' in our case. And the product of the roots formula relates the roots to the constant term. By carefully analyzing these formulas, we can extract valuable information about the roots and the coefficients. It's like deciphering a code, where each formula is a clue that leads us closer to the solution. But the real power of Vieta's formulas lies in their ability to transform the problem into a system of equations. We've essentially taken a single cubic equation and turned it into three equations involving the roots 'a' and 'd', and the unknown 'm'. This is a significant step forward, as we now have a set of tools that we can use to solve for these unknowns. The next step is to carefully simplify these equations and see how they can help us find the value of 'm'. We'll be using our algebraic skills to manipulate these equations, combining them in strategic ways to eliminate variables and isolate the unknown we're after. So, let's roll up our sleeves and get ready to dive into the algebraic manipulations that will unlock the solution.

Solving for 'a' and 'd': Simplifying the Equations

Let's simplify the equations we got from Vieta's formulas. The first equation, (a - d) + a + (a + d) = 15, simplifies beautifully to 3a = 15. This means a = 5. Awesome! We've already found one of our unknowns. This is a major breakthrough, as it significantly simplifies the remaining equations. Knowing the value of 'a' is like having a key that unlocks the next level of the problem. We can now substitute this value into the other equations, reducing the number of unknowns and making them easier to solve. Think of it like this: we've narrowed down the possibilities for the roots, and this constraint helps us to pinpoint the exact values. The third equation, (a - d) * a * (a + d) = -45, can now be written as (5 - d) * 5 * (5 + d) = -45. Dividing both sides by 5 gives us (5 - d)(5 + d) = -9. Expanding this, we get 25 - d² = -9. Rearranging, we have d² = 34, so d = ±√34. Now we have two possible values for 'd'. This means there are two possible arithmetic progressions that could be the roots of the equation. But don't worry, we'll see that both values of 'd' will lead us to the same value for 'm'. This is a common occurrence in math problems, where multiple paths can lead to the same destination. The important thing is that we've systematically narrowed down the possibilities and found the values of 'a' and 'd'. We're now in a strong position to find 'm'. We've essentially solved for the building blocks of the roots, and we're ready to assemble them into the final answer. The next step is to use these values of 'a' and 'd' in the second Vieta's formula to find 'm'. This is where the pieces of the puzzle come together, and we'll see how the relationships between the roots and coefficients lead us to the solution we've been searching for. So, let's keep our momentum going and get ready to calculate the final answer.

Finding 'm': Putting It All Together

Now, let's use the second Vieta's formula: (a - d)a + (a - d)(a + d) + a(a + d) = -m. Substitute a = 5 and d = ±√34 into this equation. Let's plug in our values. We have (5 - √34)5 + (5 - √34)(5 + √34) + 5(5 + √34) = -m. Simplifying this, we get 25 - 5√34 + (25 - 34) + 25 + 5√34 = -m. The terms with √34 cancel out, which is fantastic! This leaves us with 25 - 9 + 25 = -m, which simplifies to 41 = -m. Therefore, m = -41. We've found the value of 'm'! We've successfully navigated through the problem, using the properties of arithmetic progressions and Vieta's formulas to unlock the solution. This is a great feeling, like reaching the summit of a challenging climb. But let's just double-check to make sure we haven't made any errors along the way. We can do this by plugging our value of 'm' back into the original equation and verifying that the roots are indeed in arithmetic progression. This is a good practice to ensure the accuracy of our solution. And if we had used d = -√34, we would have arrived at the same value for 'm'. This is because the equation involves terms that are symmetric with respect to 'd', so the sign of 'd' doesn't affect the final answer. So, with confidence, we can say that we've found the correct value of 'm'. We've demonstrated the power of algebraic techniques and the beauty of mathematical connections. This problem is a testament to how seemingly abstract concepts can come together to solve concrete problems. And with that, we've conquered the Colegio Ciencias y Humanidades 5 math problem! Great job, everyone! We've cracked the code and found the value of 'm'.

Solution

Therefore, the value of m is -41.

Find the value of 'm' in the equation -x³ + 15x² + mx - 45 = 0, given that the roots are in arithmetic progression.

Solving Cubic Equations Arithmetic Progression Problem Step-by-Step