Calculate (x1 + X2)^2 Without Solving The Quadratic Equation

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Calculating (x₁ + x₂)² Without Solving the Quadratic Equation

Hey guys! Today, we're diving into a cool math problem where we need to figure out the value of $(x_1 + x_2)^2$ given a quadratic equation, but here's the catch – we're doing it without actually finding the roots $x_1$ and $x_2$. Sounds like a fun challenge, right? Let's get started!

Understanding the Quadratic Equation

First off, let's talk about the quadratic equation we're working with: $3x^2 + 7x - 2 = 0$. A quadratic equation generally looks like this: $ax^2 + bx + c = 0$, where a, b, and c are constants. In our case, a is 3, b is 7, and c is -2. The roots of a quadratic equation are the values of x that make the equation true, and these are $x_1$ and $x_2$ in our problem. Normally, you might think about using the quadratic formula to find these roots, which is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. But we're going to take a different, more elegant approach here.

Instead of directly solving for the roots, we're going to use some neat relationships between the roots and the coefficients of the quadratic equation. These relationships are super handy and can save us a lot of time and effort. So, remember those coefficients a, b, and c? They hold the key to our solution. We can express the sum and the product of the roots in terms of these coefficients. Let's dive into how this works and why it's so useful in problems like this.

Vieta's Formulas: The Key to Our Solution

Here's where Vieta's formulas come into play. These formulas give us a direct connection between the roots of a quadratic equation and its coefficients. For any quadratic equation $ax^2 + bx + c = 0$, Vieta's formulas tell us:

  1. The sum of the roots, $x_1 + x_2 = -\frac{b}{a}$
  2. The product of the roots, $x_1 \cdot x_2 = \frac{c}{a}$

These formulas are incredibly powerful because they allow us to find the sum and product of the roots without actually solving the equation. In our specific equation, $3x^2 + 7x - 2 = 0$, we have a = 3, b = 7, and c = -2. Using Vieta's formulas, we can easily find the sum of the roots:

x1+x2=−73x_1 + x_2 = -\frac{7}{3}

Now, we have the value of $x_1 + x_2$, which is the first part of what we need to calculate $(x_1 + x_2)^2$. See how much simpler this is than using the quadratic formula to find the roots individually? Vieta's formulas are a game-changer, and they're super useful for problems like this one. Next, we'll use this information to find the square of the sum, which is our final goal.

Calculating $(x_1 + x_2)^2$

Now that we know $x_1 + x_2 = -\frac{7}{3}$, calculating $(x_1 + x_2)^2$ is straightforward. We simply square the value we found:

(x1+x2)2=(−73)2(x_1 + x_2)^2 = \left(-\frac{7}{3}\right)^2

Squaring a fraction means squaring both the numerator and the denominator. So, we get:

(x1+x2)2=(−7)232=499(x_1 + x_2)^2 = \frac{(-7)^2}{3^2} = \frac{49}{9}

And there you have it! The value of $(x_1 + x_2)^2$ is $\frac{49}{9}$. We've solved the problem without ever needing to find the individual roots $x_1$ and $x_2$. This is a great example of how understanding and applying the right mathematical concepts, like Vieta's formulas, can make complex problems much simpler and more manageable.

Why This Approach Works

The beauty of this approach lies in its efficiency and elegance. Instead of grinding through the quadratic formula, which can be cumbersome and time-consuming, we used a shortcut provided by Vieta's formulas. This highlights an important aspect of problem-solving in mathematics: looking for the most direct and efficient method. Understanding the relationships between different parts of a problem, like the roots and coefficients of a quadratic equation, can often lead to quicker and simpler solutions.

This method also reduces the chance of making errors. The quadratic formula involves several steps and calculations, increasing the opportunity for mistakes. By using Vieta's formulas, we streamlined the process and minimized the risk of errors. This is particularly useful in exam situations where time and accuracy are crucial. So, remember this technique – it's a valuable tool in your math arsenal!

Expanding on the Concept

To really nail this concept, let's explore a bit further. What if we wanted to find the value of $x_1^2 + x_2^2$? We can still use Vieta's formulas, but we'll need to do a little algebraic manipulation. Remember the identity:

(x1+x2)2=x12+2x1x2+x22(x_1 + x_2)^2 = x_1^2 + 2x_1x_2 + x_2^2

We can rearrange this to get:

x12+x22=(x1+x2)2−2x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2

We already know $(x_1 + x_2)^2$ from our previous calculation, and we can find $x_1x_2$ using Vieta's formulas as well. For our equation, $3x^2 + 7x - 2 = 0$, the product of the roots is:

x1x2=ca=−23x_1x_2 = \frac{c}{a} = \frac{-2}{3}

Now we can substitute the values into our equation:

x12+x22=499−2(−23)=499+43x_1^2 + x_2^2 = \frac{49}{9} - 2\left(-\frac{2}{3}\right) = \frac{49}{9} + \frac{4}{3}

To add these fractions, we need a common denominator, which is 9. So, we convert $\frac{4}{3}$ to $\frac{12}{9}$:

x12+x22=499+129=619x_1^2 + x_2^2 = \frac{49}{9} + \frac{12}{9} = \frac{61}{9}

So, the value of $x_1^2 + x_2^2$ is $\frac{61}{9}$. This example shows how you can combine Vieta's formulas with other algebraic identities to solve even more complex problems without finding the roots directly.

Practical Applications and Real-World Relevance

You might be wondering,