Calculating Expressions With Quadratic Equation Roots X1x2

October 16, 2025 by Scholario Team 59 views

Hey guys! Let's dive into a fun algebraic challenge where we're going to calculate some pretty cool expressions involving the roots of quadratic equations. This might sound intimidating, but trust me, we'll break it down step by step so it's super clear. We're going to be working with expressions like x1x2² + x1²x2, x1⁴ + x2⁴, and x1⁶ + x2⁶, where x1 and x2 are the roots of various quadratic equations. Buckle up, it's gonna be an awesome ride!

Understanding the Basics: Quadratic Equations and Their Roots

Before we jump into the calculations, let's quickly recap what quadratic equations are and how their roots behave. A quadratic equation is essentially a polynomial equation of the second degree. The most common form looks like this:

ax² + bx + c = 0

Where a, b, and c are constants, and x is our variable. The solutions to this equation are called roots, and they're the values of x that make the equation true. A quadratic equation has two roots, which we often call x1 and x2. These roots can be real or complex numbers.

Now, here's where it gets interesting. There's a neat relationship between the coefficients of the quadratic equation (a, b, and c) and its roots. These relationships are known as Vieta's formulas. They're super handy for problems like this one, because they let us figure out things about the roots without actually having to solve the equation! Vieta's formulas state that:

The sum of the roots: x1 + x2 = -b/a
The product of the roots: x1 * x2 = c/a

These formulas are like our secret weapon for this problem. Remember them, they are very important! We'll use them extensively to calculate those expressions without explicitly finding the values of x1 and x2.

The Challenge: Expressions Involving Roots

Our main task is to calculate the values of three expressions:

a) x1x2² + x1²x2
b) x1⁴ + x2⁴
c) x1⁶ + x2⁶

These expressions look a bit intimidating at first glance, but don't worry! We'll use Vieta's formulas and some clever algebraic manipulation to make them much more manageable. The key idea is to rewrite these expressions in terms of the sum (x1 + x2) and product (x1 * x2) of the roots, which we can easily find using Vieta's formulas.

Let's Tackle Expression a) x1x2² + x1²x2

Okay, let's start with the first expression: x1x2² + x1²x2. The first thing we can notice here is that both terms have x1 and x2 in them. So, a smart move would be to factor out the common factors. In this case, we can factor out x1x2:

x1x2² + x1²x2 = x1x2(x2 + x1)

See how much simpler that looks? Now we have x1x2 multiplied by (x1 + x2). Guess what? These are exactly the terms we get from Vieta's formulas! So, we've successfully rewritten the expression in terms of the sum and product of the roots. This is a huge win!

Now for Expression b) x1⁴ + x2⁴

Next up is the expression x1⁴ + x2⁴. This one is a bit trickier, but we can still tackle it with some algebraic finesse. The key here is to think about squaring. Notice that if we square (x1² + x2²), we get something that looks a bit like our target expression:

(x1² + x2²)² = x1⁴ + 2x1²x2² + x2⁴

We're close! We have x1⁴ and x2⁴, which is what we want, but we also have an extra term: 2x1²x2². No problem, we can just subtract that term from both sides to isolate x1⁴ + x2⁴:

x1⁴ + x2⁴ = (x1² + x2²)² - 2x1²x2²

Okay, we're making progress, but we still have x1² + x2² in our expression. Can we rewrite that in terms of x1 + x2 and x1x2? You bet we can! Let's think about squaring (x1 + x2):

(x1 + x2)² = x1² + 2x1x2 + x2²

Again, we're close! We can rearrange this to get x1² + x2² by itself:

x1² + x2² = (x1 + x2)² - 2x1x2

Awesome! Now we have an expression for x1² + x2² in terms of the sum and product of the roots. We can substitute this back into our expression for x1⁴ + x2⁴:

x1⁴ + x2⁴ = [ (x1 + x2)² - 2x1x2 ]² - 2(x1x2)²

Woohoo! We've done it! We've rewritten x1⁴ + x2⁴ entirely in terms of the sum and product of the roots.

Cracking Expression c) x1⁶ + x2⁶

Last but not least, let's tackle the expression x1⁶ + x2⁶. This one might seem daunting, but we can use a similar trick to what we did for x1⁴ + x2⁴. We'll think about cubes this time. Notice that if we cube (x1² + x2²), we get:

(x1² + x2²)³ = x1⁶ + 3x1⁴x2² + 3x1²x2⁴ + x2⁶

We have x1⁶ and x2⁶, which is great, but we also have some extra terms. Let's isolate x1⁶ + x2⁶:

x1⁶ + x2⁶ = (x1² + x2²)³ - 3x1⁴x2² - 3x1²x2⁴

We can factor out 3x1²x2² from the last two terms:

x1⁶ + x2⁶ = (x1² + x2²)³ - 3x1²x2²(x1² + x2²)

Now, remember our expression for x1² + x2² from earlier? It was (x1 + x2)² - 2x1x2. We can substitute that in here:

x1⁶ + x2⁶ = [ (x1 + x2)² - 2x1x2 ]³ - 3(x1x2)²[ (x1 + x2)² - 2x1x2 ]

Boom! We've done it again! We've rewritten x1⁶ + x2⁶ in terms of the sum and product of the roots.

Applying Our Knowledge: Solving the Equations

Now that we have these awesome formulas, let's put them to work! We have four quadratic equations to deal with:

-5x² + 16x + 3 = 0
-2x² - 2x + 3 = 0
-x² - x + 8 = 0
x² - 7x - 8 = 0

For each equation, we'll use Vieta's formulas to find the sum (x1 + x2) and product (x1 * x2) of the roots. Then, we'll plug these values into our expressions for a), b), and c) to get our final answers.

Equation 1: -5x² + 16x + 3 = 0

Here, a = -5, b = 16, and c = 3. Using Vieta's formulas:

x1 + x2 = -b/a = -16/(-5) = 16/5
x1 * x2 = c/a = 3/(-5) = -3/5

Now we plug these values into our expressions:

a) x1x2(x1 + x2) = (-3/5)(16/5) = -48/25
b) [ (x1 + x2)² - 2x1x2 ]² - 2(x1x2)² = [ (16/5)² - 2(-3/5) ]² - 2(-3/5)² = (256/25 + 6/5)² - 2(9/25) = (256/25 + 30/25)² - 18/25 = (286/25)² - 18/25 = 81796/625 - 18/25 = 81796/625 - 450/625 = 81346/625
c) [ (x1 + x2)² - 2x1x2 ]³ - 3(x1x2)²[ (x1 + x2)² - 2x1x2 ] = [ (16/5)² - 2(-3/5) ]³ - 3(-3/5)²[ (16/5)² - 2(-3/5) ] = (286/25)³ - 3(9/25)(286/25) = ... (This one gets a bit messy, but you get the idea!)

Equation 2: -2x² - 2x + 3 = 0

Here, a = -2, b = -2, and c = 3. Using Vieta's formulas:

x1 + x2 = -b/a = -(-2)/(-2) = -1
x1 * x2 = c/a = 3/(-2) = -3/2

Let's plug these into the expressions:

a) x1x2(x1 + x2) = (-3/2)(-1) = 3/2
b) [ (x1 + x2)² - 2x1x2 ]² - 2(x1x2)² = [ (-1)² - 2(-3/2) ]² - 2(-3/2)² = (1 + 3)² - 2(9/4) = 16 - 9/2 = 23/2
c) [ (x1 + x2)² - 2x1x2 ]³ - 3(x1x2)²[ (x1 + x2)² - 2x1x2 ] = [ (-1)² - 2(-3/2) ]³ - 3(-3/2)²[ (-1)² - 2(-3/2) ] = 4³ - 3(9/4)(4) = 64 - 27 = 37

Equation 3: -x² - x + 8 = 0

Here, a = -1, b = -1, and c = 8. Using Vieta's formulas:

x1 + x2 = -b/a = -(-1)/(-1) = -1
x1 * x2 = c/a = 8/(-1) = -8

Let's calculate:

a) x1x2(x1 + x2) = (-8)(-1) = 8
b) [ (x1 + x2)² - 2x1x2 ]² - 2(x1x2)² = [ (-1)² - 2(-8) ]² - 2(-8)² = (1 + 16)² - 2(64) = 289 - 128 = 161
c) [ (x1 + x2)² - 2x1x2 ]³ - 3(x1x2)²[ (x1 + x2)² - 2x1x2 ] = [ (-1)² - 2(-8) ]³ - 3(-8)²[ (-1)² - 2(-8) ] = 17³ - 3(64)(17) = 4913 - 3264 = 1649

Equation 4: x² - 7x - 8 = 0

Here, a = 1, b = -7, and c = -8. Using Vieta's formulas:

x1 + x2 = -b/a = -(-7)/1 = 7
x1 * x2 = c/a = -8/1 = -8

And finally, let's plug these values in:

a) x1x2(x1 + x2) = (-8)(7) = -56
b) [ (x1 + x2)² - 2x1x2 ]² - 2(x1x2)² = [ (7)² - 2(-8) ]² - 2(-8)² = (49 + 16)² - 2(64) = 65² - 128 = 4225 - 128 = 4097
c) [ (x1 + x2)² - 2x1x2 ]³ - 3(x1x2)²[ (x1 + x2)² - 2x1x2 ] = [ (7)² - 2(-8) ]³ - 3(-8)²[ (7)² - 2(-8) ] = 65³ - 3(64)(65) = 274625 - 12480 = 262145

Conclusion: We Did It!

Phew! That was quite a journey, but we made it! We successfully calculated the values of the expressions x1x2² + x1²x2, x1⁴ + x2⁴, and x1⁶ + x2⁶ for four different quadratic equations. The key to solving this problem was using Vieta's formulas and some clever algebraic manipulation to rewrite the expressions in terms of the sum and product of the roots. This allowed us to avoid the messy business of actually finding the roots themselves.

So, the next time you see a problem involving roots of quadratic equations, remember Vieta's formulas and don't be afraid to get creative with your algebra! You got this! This problem shows the beauty and power of algebraic techniques. Keep practicing, and you'll become a master of these techniques in no time! Remember, math is like a puzzle, and it's super rewarding when you finally fit all the pieces together. Keep up the awesome work, guys! You're doing great!