Solve (x-7)(x+9)=0: Find Equivalent Equations

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of algebra, specifically tackling equations. You know, those things that look like a puzzle but are actually super useful in tons of cool ways. We've got a question that's a classic: we're given an equation in factored form, $(x-7)(x+9)=0$, and we need to find which of the provided multiple-choice options has exactly the same solutions. This is a fundamental skill, and understanding it will make tackling more complex math problems a breeze. Think of it like finding the secret twin of an equation – it might look different on the outside, but it behaves exactly the same on the inside. We're going to break down how to go from factored form to standard quadratic form and, more importantly, why it works. So, grab your notebooks, maybe a snack, and let's get this math party started! We'll cover the zero product property, how to expand binomials, and how to connect the dots between different forms of quadratic equations. This isn't just about getting the right answer; it's about understanding the process so you can solve any similar problem that comes your way. Ready to unlock the secrets of equivalent equations? Let's go!

Understanding the Zero Product Property: The Key to Our Solutions

The question asks us to find an equation that has the exact same solutions as $(x-7)(x+9)=0$. The most crucial concept we need to understand here is the Zero Product Property. This property is the bedrock of solving polynomial equations, especially when they are presented in a factored form like ours. Simply put, the Zero Product Property states that if the product of two or more factors is zero, then at least one of those factors must be zero. So, for our equation, $(x-7)(x+9)=0$, this means that either the first factor, $(x-7)$, equals zero, OR the second factor, $(x+9)$, equals zero. This is the magic that allows us to find the individual solutions. Let's break it down. If $(x-7) = 0$, we can easily solve for $x$ by adding 7 to both sides, giving us $x=7$. Similarly, if $(x+9) = 0$, we subtract 9 from both sides to get $x=-9$. So, the solutions to the equation $(x-7)(x+9)=0$ are $x=7$ and $x=-9$. Now, our mission is to find which of the given options, when solved, yields these exact same two values for $x$. This means we're not just looking for any equation, but one that is mathematically equivalent to our original. The Zero Product Property is our primary tool, and recognizing its application is step one in mastering equation solving. It's super powerful because it allows us to transform a single, complex-looking equation into two simpler, linear equations that are easy to solve. So, whenever you see a product of terms set equal to zero, immediately think: "Ah, Zero Product Property time!"

Expanding the Binomials: Transforming to Standard Form

Since our original equation $(x-7)(x+9)=0$ is in factored form, and the answer choices are in standard quadratic form ($ax^2+bx+c=0$), we need to convert our equation into that standard form. This involves expanding the binomials. Don't worry, it's not as scary as it sounds! We're essentially multiplying the two expressions in the parentheses together. The most common method for this is often called FOIL (First, Outer, Inner, Last), which is just a mnemonic to remember all the pairs you need to multiply. Let's apply it to $(x-7)(x+9)$:

• First: Multiply the first terms in each binomial: $x * x = x^2$
• Outer: Multiply the outer terms: $x * 9 = 9x$
• Inner: Multiply the inner terms: $-7 * x = -7x$
• Last: Multiply the last terms: $-7 * 9 = -63$

Now, we combine these results: $x^2 + 9x - 7x - 63$. The next step is to combine like terms. Here, $9x$ and $-7x$ are like terms because they both have the variable $x$ raised to the power of 1. Combining them gives us $2x$. So, our expanded equation becomes $x^2 + 2x - 63$. Since our original equation was set equal to zero, our expanded form is also equal to zero. Therefore, $x^2 + 2x - 63 = 0$. This is the standard quadratic form of our original equation. Now, we can compare this directly to the answer choices. The goal of expanding is to get the polynomial into a recognizable form, which in this case is the standard form $ax^2+bx+c=0$. Mastering this expansion technique is key because many algebra problems are presented this way, and knowing how to revert to or arrive at the factored form (or vice versa) is super handy. Remember, FOIL is just one way; the distributive property is the underlying principle. You're essentially distributing each term of the first binomial to each term of the second binomial. Keep practicing this, and it'll become second nature!

Matching the Solutions: Finding the Equivalent Equation

Alright guys, we've done the heavy lifting! We started with $(x-7)(x+9)=0$ and used the Zero Product Property to find its solutions: $x=7$ and $x=-9$. Then, we expanded the binomials to convert the equation into standard quadratic form, arriving at $x^2 + 2x - 63 = 0$. Now comes the final, satisfying step: matching this standard form equation with one of the options provided. Let's look at our options:

• A. $x^2 - 2x - 63 = 0$
• B. $x^2 + 2x - 63 = 0$
• C. $x^2 - 2x + 63 = 0$
• D. $x^2 + 2x + 63 = 0$

Compare the equation we derived, $x^2 + 2x - 63 = 0$, with each option. We can see a direct match with option B. Both equations have the same $x^2$ term, the same $x$ term (with coefficient +2), and the same constant term (-63). This means they represent the exact same relationship between $x$ and the resulting value of zero, and therefore, they must have the exact same solutions. The beauty of equivalent equations is that they can be written in different forms, but their core solutions remain unchanged. This is why understanding how to expand (from factored to standard form) and factor (from standard to factored form) is so critical in algebra. It allows us to move between different representations of the same mathematical idea. So, the equation that has exactly the same solutions as $(x-7)(x+9)=0$ is B. $x^2 + 2x - 63 = 0$. You totally nailed it if you got this! It's all about connecting the dots between the factored form and the standard form, ensuring you perform the algebraic manipulations correctly.

Why the Other Options Are Incorrect: A Quick Check

It's always a good idea to quickly check why the other options don't work, guys. This reinforces our understanding and helps catch any potential mistakes. We found that $(x-7)(x+9)=0$ expands to $x^2 + 2x - 63 = 0$. Let's see what happens if we try to factor the other options or consider their potential solutions:

• Option A: $x^2 - 2x - 63 = 0$ If we were to factor this, we'd look for two numbers that multiply to -63 and add up to -2. Those numbers are -9 and +7. So, this equation factors to $(x-9)(x+7)=0$. The solutions would be $x=9$ and $x=-7$. These are not the same solutions as our original equation ($x=7, x=-9$). Notice the signs are flipped on both the $x$ term and the solutions, which is a common mistake. This confirms Option A is incorrect.

• Option C: $x^2 - 2x + 63 = 0$ Here, we need two numbers that multiply to +63 and add up to -2. It's tricky to find integers that do this. If we try the quadratic formula, $x = [-b ± \sqrt{b^2-4ac}] / 2a$, with $a=1, b=-2, c=63$, we get $x = [2 ± \sqrt{(-2)^2 - 4(1)(63)}] / 2(1) = [2 ± \sqrt{4 - 252}] / 2 = [2 ± \sqrt{-248}] / 2$. Since the discriminant ($b^2-4ac$) is negative, this equation has no real solutions. Our original equation definitely has real solutions ($x=7, x=-9$), so this cannot be equivalent.

• Option D: $x^2 + 2x + 63 = 0$ Similarly, for this equation, we need two numbers that multiply to +63 and add up to +2. Again, finding integer pairs is difficult. Using the quadratic formula with $a=1, b=2, c=63$, we get $x = [-2 ± \sqrt{(2)^2 - 4(1)(63)}] / 2(1) = [-2 ± \sqrt{4 - 252}] / 2 = [-2 ± \sqrt{-248}] / 2$. The discriminant is again negative, meaning no real solutions. This is also not equivalent to our original equation.

By examining the other options, we can confidently confirm that Option B is the only one that shares the exact same solutions as $(x-7)(x+9)=0$. It's all about precision in algebra – a single sign change can lead to entirely different outcomes, so always double-check your work!

Conclusion: Mastering Equivalent Quadratic Equations

So there you have it, folks! We've successfully navigated the world of quadratic equations, starting from a factored form, $(x-7)(x+9)=0$, and pinpointing its equivalent in standard form. The key takeaways are the Zero Product Property, which allows us to find the solutions from the factored form ($x=7$ and $x=-9$), and the process of expanding binomials (using FOIL or the distributive property) to convert the equation into the standard quadratic form $ax^2+bx+c=0$. By expanding $(x-7)(x+9)$, we found it equals $x^2 + 2x - 63$. Comparing this to the multiple-choice options, we definitively identified B. $x^2 + 2x - 63 = 0$ as the equation with exactly the same solutions. We also took the time to analyze why the other options were incorrect, reinforcing the importance of paying attention to signs and understanding the nature of solutions (real vs. complex). This skill of recognizing and manipulating equivalent equations is super valuable, not just for test-taking, but for understanding the deeper connections in algebra. Whether you're solving equations, graphing parabolas, or working with functions, being able to move between different forms of a quadratic equation will make you a math superhero! Keep practicing, keep asking questions, and remember that every problem you solve builds your confidence and your understanding. That’s all for today’s math breakdown here at Plastik Magazine. Stay curious, stay awesome, and we’ll catch you in the next one!