Solving Quadratic Equations And Word Problems A Step-by-Step Guide

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Hey guys! Today, we're diving deep into the fascinating world of quadratic equations and word problems. We'll break down some examples step by step, making sure you grasp the core concepts. Whether you're tackling homework or just brushing up on your math skills, this guide is for you. Let's get started!

1. Solving the Quadratic Equation: 9x² + 9x + 2 = 0

When it comes to solving quadratic equations, one of the most common methods is factoring. Factoring allows us to break down a complex quadratic expression into simpler components, making it easier to find the roots or solutions. Let's jump into our first example, which is the quadratic equation 9x² + 9x + 2 = 0.

Understanding the Quadratic Equation

First off, let's make sure we're all on the same page. A quadratic equation is generally expressed in the form ax² + bx + c = 0, where a, b, and c are constants, and x is the variable we're trying to solve for. In our case, a = 9, b = 9, and c = 2. The goal here is to find the values of x that satisfy this equation. These values are also known as the roots or solutions of the equation.

Factoring the Quadratic Equation

Now, let's get to the fun part – factoring! The key to factoring a quadratic equation is to find two binomials that, when multiplied together, give us the original quadratic expression. For 9x² + 9x + 2, we need to find two binomials of the form (Ax + B)(Cx + D) such that:

  • A * C = 9
  • B * D = 2
  • (A * D) + (B * C) = 9

This might sound a bit daunting, but it's really a puzzle. We need to find the right combination of numbers that fit these conditions.

After some trial and error (or using factoring techniques), we find that the quadratic expression can be factored as:

(3x + 1)(3x + 2) = 0

Finding the Roots

Great! We've factored the equation. Now, to find the roots, we set each factor equal to zero:

  1. 3x + 1 = 0
  2. 3x + 2 = 0

Solving the first equation (3x + 1 = 0) gives us:

3x = -1 x = -1/3

Solving the second equation (3x + 2 = 0) gives us:

3x = -2 x = -2/3

So, the solutions to the quadratic equation 9x² + 9x + 2 = 0 are x = -1/3 and x = -2/3.

Verifying the Solutions

It's always a good idea to verify our solutions. We can do this by plugging the values of x back into the original equation to see if they satisfy it.

For x = -1/3:

9(-1/3)² + 9(-1/3) + 2 = 9(1/9) - 3 + 2 = 1 - 3 + 2 = 0

For x = -2/3:

9(-2/3)² + 9(-2/3) + 2 = 9(4/9) - 6 + 2 = 4 - 6 + 2 = 0

Both solutions check out! So, we've successfully solved the quadratic equation by factoring. Factoring can seem tricky at first, but with practice, you'll get the hang of it. Remember, the key is to break down the problem into smaller parts and think step by step. Whether you're solving for x in a simple equation or tackling more complex problems, understanding the basics of factoring is super important.

2. Dealing with the Quartic Equation: 2x⁴ - 5x - 3 = 0

Alright, let's crank up the complexity a notch. We're now facing a quartic equation: 2x⁴ - 5x - 3 = 0. Unlike our earlier quadratic, this one's got a term with x raised to the power of 4, making it a bit of a beast. But don't worry, we'll tame it together. Solving quartic equations can be tricky, and sometimes there isn’t a straightforward method like factoring a simple quadratic. However, we can explore a few strategies that might help us crack this nut.

Understanding Quartic Equations

First off, let’s get a clear picture of what we’re dealing with. A quartic equation is a polynomial equation of the fourth degree. The general form looks something like ax⁴ + bx³ + cx² + dx + e = 0, where a, b, c, d, and e are constants, and a isn't zero (otherwise, it wouldn't be a quartic equation!). Our specific equation is 2x⁴ - 5x - 3 = 0. Notice that some terms (like the x³ and x² terms) are missing, which actually simplifies things a little.

Strategies for Solving

Given the complexity of quartics, there’s no single, foolproof method. We often need to use a combination of techniques. Here are a few strategies we can consider:

  1. Rational Root Theorem: This theorem helps us find potential rational roots (roots that can be expressed as fractions). It states that if a rational number p/q is a root of the polynomial, then p must be a factor of the constant term, and q must be a factor of the leading coefficient. In our case, the constant term is -3 and the leading coefficient is 2. So, potential rational roots are ±1, ±3, ±1/2, and ±3/2.

  2. Synthetic Division: Once we have potential roots, we can use synthetic division to test them. If a number is a root, synthetic division will give us a remainder of zero. This also helps us reduce the degree of the polynomial.

  3. Factoring by Grouping: Sometimes, we can rearrange and group terms in a way that allows us to factor by grouping. This is more likely to work if there’s a clear pattern or common factors among the terms.

  4. Numerical Methods: For equations that are particularly tough to solve algebraically, we can resort to numerical methods like the Newton-Raphson method or using computational tools to approximate the roots.

Applying the Rational Root Theorem

Let’s start with the Rational Root Theorem. Our potential rational roots are ±1, ±3, ±1/2, and ±3/2. We can test these one by one using synthetic division or direct substitution.

Let’s try x = -1:

2(-1)⁴ - 5(-1) - 3 = 2(1) + 5 - 3 = 2 + 5 - 3 = 4

So, x = -1 is not a root.

Let’s try x = 1:

2(1)⁴ - 5(1) - 3 = 2 - 5 - 3 = -6

So, x = 1 is not a root either.

Let's try x = -3/2:

This might look intimidating, but let's break it down:

2(-3/2)⁴ - 5(-3/2) - 3 = 2(81/16) + 15/2 - 3 = 81/8 + 15/2 - 3

To combine these, we need a common denominator, which is 8:

81/8 + 60/8 - 24/8 = (81 + 60 - 24) / 8 = 117 / 8

So, x = -3/2 is not a root.

Let's try x = 3/2:

2(3/2)⁴ - 5(3/2) - 3 = 2(81/16) - 15/2 - 3 = 81/8 - 15/2 - 3

Again, let's get a common denominator of 8:

81/8 - 60/8 - 24/8 = (81 - 60 - 24) / 8 = -3 / 8

So, x = 3/2 is not a root.

Trying Other Methods

Since the Rational Root Theorem didn’t immediately give us a root, we might consider numerical methods or factoring by grouping. Factoring by grouping doesn't seem straightforward here, so let’s think about numerical methods. In a real-world scenario, we might use a calculator or software to find approximate roots. These tools employ algorithms to find the values of x that make the equation equal to zero.

Alternatively, we can try to rewrite the equation to make it easier to handle. For example, we could look for substitutions that simplify the equation. This approach requires a bit of algebraic intuition and might not always be obvious.

The Challenge of Quartic Equations

This example illustrates a crucial point: quartic equations can be tough! They often require a combination of strategies and might not have simple, rational solutions. If you’re faced with a quartic equation in an exam, it’s worth spending some time trying to find rational roots, but don’t be afraid to use numerical methods or software tools if allowed.

In conclusion, while we didn't find a simple solution for 2x⁴ - 5x - 3 = 0 using elementary methods, we explored various strategies. Remember, solving complex equations is often about trying different approaches and using the tools at your disposal. Whether you're dealing with quartics, cubics, or other polynomials, persistence and a methodical approach are key.

3. Solving Word Problems: Finding Two Numbers

Now, let's switch gears and tackle a classic type of word problem. Word problems often seem daunting because they require us to translate real-world scenarios into mathematical equations. But don't worry, we'll break it down. The problem we're going to solve is: “The sum of two numbers is 22, and their product is 117. Find these numbers.” These problems are a fantastic way to apply math in real life, and once you get the hang of them, they're really not that bad!

Translating Words into Equations

The first step in solving any word problem is to translate the given information into mathematical equations. This might seem tricky, but it's really about identifying the key pieces of information and expressing them in mathematical terms.

In our problem, we have two pieces of information:

  1. The sum of two numbers is 22.
  2. The product of the same two numbers is 117.

Let's call our two numbers x and y. Now we can translate these statements into equations:

  1. x + y = 22
  2. x * y = 117

Solving the System of Equations

Now we have a system of two equations with two variables. There are a few methods we can use to solve this system, such as substitution or elimination. In this case, substitution seems like a good approach. We can solve the first equation for one variable (let's solve for y) and then substitute that expression into the second equation.

From the first equation (x + y = 22), we can solve for y:

y = 22 - x

Now, substitute this expression for y into the second equation (x * y = 117):

x * (22 - x) = 117

Forming a Quadratic Equation

Expanding the equation, we get:

22x - x² = 117

Rearranging the terms to form a quadratic equation, we have:

x² - 22x + 117 = 0

Now we have a quadratic equation that we can solve. We can use factoring, the quadratic formula, or completing the square. Factoring seems like the easiest approach here. We need to find two numbers that multiply to 117 and add up to -22. These numbers are -13 and -9.

So, we can factor the quadratic equation as:

(x - 13)(x - 9) = 0

Finding the Solutions

Setting each factor equal to zero gives us the solutions for x:

  1. x - 13 = 0 => x = 13
  2. x - 9 = 0 => x = 9

Now, we can find the corresponding values for y using the equation y = 22 - x:

If x = 13: y = 22 - 13 = 9

If x = 9: y = 22 - 9 = 13

So, the two numbers are 9 and 13. It doesn't matter which we call x and which we call y, as long as we have both numbers.

Verifying the Solutions

It's always a good idea to verify our solutions in the original problem. Let's check:

  1. Sum: 13 + 9 = 22 (Correct!)
  2. Product: 13 * 9 = 117 (Correct!)

Key Takeaways

Word problems can be challenging, but they become much more manageable when you break them down into steps. Here are the key takeaways:

  1. Translate the words into equations.
  2. Solve the system of equations.
  3. Verify your solutions.

Whether it's finding two numbers, calculating distances, or solving mixture problems, the same approach applies. Practice is the key, so keep tackling those word problems!

Wrapping Up

So there you have it, guys! We've tackled quadratic equations, a tricky quartic equation, and a classic word problem. We've explored different techniques like factoring, using the Rational Root Theorem, and translating real-world scenarios into mathematical equations. Remember, math is like a muscle – the more you use it, the stronger it gets. Keep practicing, stay curious, and you'll be solving equations like a pro in no time! Whether you're mastering quadratic equations or conquering word problems, the key is to break things down, stay organized, and never be afraid to try different approaches. Happy solving!