Finding Sum Of Squares Of Roots: Polynomial Equation

Oct 17, 2025 by SLV Team 53 views

Hey guys! Today, we're diving into a cool problem involving polynomial equations and their roots. Specifically, we're going to figure out how to find the sum of the squares of the roots of a cubic polynomial. Sounds like a mouthful, right? But trust me, we'll break it down and make it super easy to understand. Let's get started!

Understanding the Problem

So, here’s the deal. We have a cubic polynomial equation: $2x^3 - 4x^2 + 6x - 10 = 0$ . This equation has three roots, which we'll call $x_1$ , $x_2$ , and $x_3$ . Our mission, should we choose to accept it (and we do!), is to find the value of $x_1^2 + x_2^2 + x_3^2$ . In simpler terms, we need to find the sum of the squares of these roots. Now, you might be thinking, “Do I need to find the roots individually and then square them?” Nope! There's a much slicker way to do this using some neat relationships between the roots and coefficients of the polynomial. We're going to use Vieta's formulas and a little bit of algebraic magic to solve this. This approach is not only faster but also gives us a deeper understanding of how polynomials behave. Remember, math isn't just about finding the answer; it's about understanding the process. Think of it like this: if you know how to bake a cake, you can bake any cake! So, let's get into the nitty-gritty and see how this works. We'll start by refreshing our memory on Vieta's formulas, because they are the real MVPs in this scenario. Trust me, by the end of this, you'll be looking at polynomial equations in a whole new light.

Vieta's Formulas: The Key to Our Solution

Okay, let's talk about Vieta's formulas. These formulas are like secret codes that link the coefficients of a polynomial to the sums and products of its roots. They are super handy for problems like this one, where we need to find relationships between the roots without actually solving for them individually. For a general cubic polynomial equation of the form $ax^3 + bx^2 + cx + d = 0$ , Vieta's formulas tell us the following:

The sum of the roots: $x_1 + x_2 + x_3 = -\frac{b}{a}$
The sum of the products of the roots taken two at a time: $x_1x_2 + x_1x_3 + x_2x_3 = \frac{c}{a}$
The product of the roots: $x_1x_2x_3 = -\frac{d}{a}$

Now, let's apply these formulas to our specific polynomial, which is $2x^3 - 4x^2 + 6x - 10 = 0$ . Here, $a = 2$ , $b = -4$ , $c = 6$ , and $d = -10$ . Using Vieta's formulas, we get:

$x_1 + x_2 + x_3 = -\frac{-4}{2} = 2$
$x_1x_2 + x_1x_3 + x_2x_3 = \frac{6}{2} = 3$
$x_1x_2x_3 = -\frac{-10}{2} = 5$

These three equations are our foundation. We now have the sum of the roots, the sum of the products of the roots taken two at a time, and the product of all three roots. But remember, our goal is to find $x_1^2 + x_2^2 + x_3^2$ . How do we get there from here? Well, we need to do a little algebraic maneuvering. We're going to use a clever trick that involves squaring the first equation and then subtracting a multiple of the second equation. This will help us isolate the expression we're looking for. It's like a mathematical puzzle, and we're about to fit the pieces together. So, stick with me, and let's see how this algebraic magic works!

The Algebraic Trick: Squaring and Subtracting

Alright, so we've got Vieta's formulas giving us some valuable information. We know that $(x_1 + x_2 + x_3) = 2$ and $(x_1x_2 + x_1x_3 + x_2x_3) = 3$ . Now, we need to find a way to relate these to $x_1^2 + x_2^2 + x_3^2$ . Here’s the trick: let’s square the first equation, $(x_1 + x_2 + x_3) = 2$ . When we do that, we get:

$(x_1 + x_2 + x_3)^2 = 2^2$

Expanding the left side, we have:

$x_1^2 + x_2^2 + x_3^2 + 2(x_1x_2 + x_1x_3 + x_2x_3) = 4$

Notice something cool here? We've got the $x_1^2 + x_2^2 + x_3^2$ term that we're after! We also have the term $2(x_1x_2 + x_1x_3 + x_2x_3)$ , and guess what? We know the value of $(x_1x_2 + x_1x_3 + x_2x_3)$ from Vieta's formulas—it's 3. So, we can substitute that value into the equation:

$x_1^2 + x_2^2 + x_3^2 + 2(3) = 4$

Now, it’s just a matter of simplifying and solving for $x_1^2 + x_2^2 + x_3^2$ . We have:

$x_1^2 + x_2^2 + x_3^2 + 6 = 4$

Subtracting 6 from both sides, we get:

$x_1^2 + x_2^2 + x_3^2 = 4 - 6$

$x_1^2 + x_2^2 + x_3^2 = -2$

And there you have it! We’ve found that the value of $x_1^2 + x_2^2 + x_3^2$ is -2. This little algebraic trick is super useful in many polynomial problems. By squaring the sum of the roots and using the sum of the products of the roots, we were able to isolate the expression we needed. It's like unlocking a secret level in a game! Now that we've solved the problem, let's recap the steps to make sure we've got it all down pat.

Recap and Final Answer

Okay, let's take a step back and quickly recap what we've done to solve this problem. This will help solidify the method in your mind so you can tackle similar questions in the future. First, we identified the cubic polynomial equation: $2x^3 - 4x^2 + 6x - 10 = 0$ . We knew we needed to find the value of $x_1^2 + x_2^2 + x_3^2$ , where $x_1$ , $x_2$ , and $x_3$ are the roots of the equation.

Then, we brought in the big guns: Vieta's formulas. These formulas allowed us to relate the coefficients of the polynomial to the sums and products of its roots. We found:

$x_1 + x_2 + x_3 = 2$
$x_1x_2 + x_1x_3 + x_2x_3 = 3$
$x_1x_2x_3 = 5$

Next, we used a clever algebraic trick. We squared the sum of the roots, $(x_1 + x_2 + x_3)^2$ , which gave us:

$x_1^2 + x_2^2 + x_3^2 + 2(x_1x_2 + x_1x_3 + x_2x_3) = 4$

We then substituted the value of $(x_1x_2 + x_1x_3 + x_2x_3)$ from Vieta's formulas into this equation:

$x_1^2 + x_2^2 + x_3^2 + 2(3) = 4$

Simplifying, we got:

$x_1^2 + x_2^2 + x_3^2 + 6 = 4$

Finally, we solved for $x_1^2 + x_2^2 + x_3^2$ :

$x_1^2 + x_2^2 + x_3^2 = 4 - 6 = -2$

So, the value of $x_1^2 + x_2^2 + x_3^2$ is -2. Therefore, the correct answer is D. -2. Isn't it amazing how we can find these relationships without actually finding the individual roots? Vieta's formulas and a bit of algebraic manipulation can take us a long way! Now, you’ve got another tool in your math toolkit. Go forth and conquer more polynomial problems! You've got this!