Solving (x-27)(x-10)=60: Are 10 & 30 The Right Answers?

by Andrew McMorgan 56 views

Hey there, Plastik Magazine fam! Ever stared at a quadratic equation and wondered if a specific number actually fits? It's like trying on shoes – some fit perfectly, others just don't! Today, we're diving deep into a fascinating problem: determining whether the values of 10 and 30 are actual solutions to the quadratic equation (x-27)(x-10)=60. This isn't just about math; it's about understanding the mechanics of equations and how to verify if a given number truly makes the equation true. So grab your calculators, or just your thinking caps, because we're about to demystify this mathematical puzzle together. We'll break down the process step-by-step, ensuring you not only get the answer but also understand the "why" behind it. Get ready to flex those brain muscles, guys!

Understanding the World of Quadratic Equations

Alright, guys, let's kick things off by getting cozy with quadratic equations. These aren't just fancy mathematical terms; they're super common in the real world, modeling everything from the trajectory of a thrown ball to the optimal pricing for a product. At its heart, a quadratic equation is any equation that can be rearranged into the standard form: ax² + bx + c = 0, where 'a', 'b', and 'c' are constants, and 'a' cannot be zero. The key giveaway is that x is raised to the power of two (that's the "quad" part, meaning square). When you see an x² in an equation, you're officially in quadratic territory!

Think of solutions or roots to a quadratic equation as the specific values for x that make the entire equation true. If you plug in a solution, both sides of the equals sign will balance out perfectly. Because of that x² term, most quadratic equations have two distinct solutions, though sometimes they might have one repeated solution or even no real solutions at all. In our case, we're working with (x-27)(x-10)=60, which might not look like the standard form ax² + bx + c = 0 right away. But trust me, if we expanded it, we'd absolutely find an x² term hiding in there. We'd multiply x by x to get x², then x by -10, -27 by x, and -27 by -10. That would give us x² - 10x - 27x + 270 = 60, which simplifies to x² - 37x + 270 = 60. If we then subtract 60 from both sides, we get x² - 37x + 210 = 0. Voila! Standard quadratic form. Understanding this foundational concept of what a quadratic equation is and what its solutions represent is absolutely crucial before we jump into testing our specific values. It's the bedrock of our entire investigation into whether 10 and 30 are indeed the elusive answers we're looking for. Knowing this makes the process of substituting values and verifying solutions much more meaningful. So, now that we're all on the same page about the basics, let's get to the fun part: putting our potential solutions to the test!

The Challenge: Testing Potential Solutions for (x-27)(x-10)=60

Alright, team, now that we're clear on what quadratic equations are all about, it's time to tackle the main event: testing potential solutions for our specific equation, (x-27)(x-10)=60. This isn't a complex, mind-bending task; it's actually pretty straightforward, almost like a forensic investigation for numbers! The core idea is simple: if a value is a solution to an equation, then when you substitute that value in for x, the equation should hold true. That means the left side of the equation should exactly equal the right side. If it doesn't, then that value isn't a solution. Easy, right?

We've been given two specific numbers to check: 10 and 30. Our mission, should we choose to accept it, is to take each of these numbers, one at a time, plug them into the x spots in our equation (x-27)(x-10)=60, and then do the math. We'll meticulously calculate the value of the left-hand side (LHS) of the equation and see if it matches the right-hand side (RHS), which in this case is 60. This process of substituting values is your go-to method for verifying solutions in any algebraic equation, not just quadratics. It's a fundamental skill that will serve you well in all your mathematical adventures. There’s no need to guess or assume; we're going to use solid, undeniable arithmetic to get our answers. We're not trying to solve the equation from scratch here – that's a whole other adventure we can explore later. For now, our focus is purely on checking the given values. This approach allows us to confirm or deny the validity of 10 and 30 as solutions without needing to factor, use the quadratic formula, or complete the square. It’s a direct and efficient way to verify if these numbers are indeed the roots we're looking for. So, let's roll up our sleeves and get ready to plug and chug! We'll start with x = 10 and see if it stands up to the mathematical scrutiny.

Is x = 10 a Solution? Let's Find Out!

Okay, guys, first up in our solution verification challenge is the number 10. We're going to take x = 10 and carefully substitute it into our quadratic equation: (x-27)(x-10)=60. This is where the magic (or lack thereof) happens!

Let's break it down:

  1. Substitute x = 10 into the equation: Our equation is (x-27)(x-10) = 60. Replacing every 'x' with '10', we get: (10 - 27)(10 - 10) = 60.

  2. Calculate the expressions inside the parentheses: For the first set of parentheses: 10 - 27 = -17. For the second set of parentheses: 10 - 10 = 0.

  3. Multiply the results from the parentheses: Now we have: (-17) * (0) = 60.

  4. Perform the multiplication: -17 multiplied by 0 is, without a doubt, 0.

  5. Compare the result to the right-hand side of the equation: So, our calculation gives us 0 = 60.

Now, pause and look at that last line: 0 = 60. Is that a true statement? Absolutely not! Zero does not equal sixty. This means that when we plugged in x = 10, the left side of the equation did not balance with the right side. Therefore, we can definitively conclude that x = 10 is NOT a solution to the quadratic equation (x-27)(x-10)=60.

This is a crucial point in solution verification. Even if one part of the equation simplifies nicely, it's the final comparison that determines the outcome. The fact that the second parenthesis 10 - 10 became 0 was a pretty quick way to resolve the left side to zero, but the outcome would be the same no matter what values were inside the first parenthesis. This methodical approach ensures we don't jump to conclusions. We followed the rules of algebra, performed the substitution, and evaluated the expression. The numbers spoke for themselves, telling us clearly that x = 10 doesn't satisfy the conditions of this specific quadratic equation. So, one down, one to go! Let's see if x = 30 fares any better in our mathematical challenge.

How About x = 30? Testing the Second Value

Alright, Plastik crew, we've seen that x = 10 didn't quite cut it. Now it's time to give the second potential solution a shot: 30. We're going to repeat the exact same process, substituting x = 30 into our equation (x-27)(x-10)=60 and seeing if this number can truly make the equation sing. This testing values method is consistent and reliable, so let's stick to our step-by-step approach!

Here’s how we break down the verification for x = 30:

  1. Substitute x = 30 into the equation: Our equation is (x-27)(x-10) = 60. Replacing every 'x' with '30', we now have: (30 - 27)(30 - 10) = 60.

  2. Calculate the expressions inside the parentheses: For the first set of parentheses: 30 - 27 = 3. For the second set of parentheses: 30 - 10 = 20.

  3. Multiply the results from the parentheses: Now we have: (3) * (20) = 60.

  4. Perform the multiplication: 3 multiplied by 20 gives us 60.

  5. Compare the result to the right-hand side of the equation: So, our calculation yields 60 = 60.

Aha! Now that's what we're talking about! Look at that last line: 60 = 60. Is that a true statement? Absolutely! The left side of the equation perfectly matches the right side. This means that when we plugged in x = 30, the equation held true; it balanced out perfectly. Therefore, we can confidently conclude that x = 30 IS a solution to the quadratic equation (x-27)(x-10)=60.

This is a fantastic example of how to verify solutions in a practical way. The step-by-step process of substituting values and meticulously calculating each step prevents errors and leaves no room for doubt. Unlike x = 10, which led to a false statement, x = 30 successfully passed the test, making it a genuine root of this specific quadratic equation. This brings us to a crucial understanding of what it means to be a solution. It's not just a number that might work; it's a number that definitely makes the equation true. Our investigation is shaping up, and we're building a clear picture of which values make the cut.

What Does This Mean? Interpreting Our Findings

So, guys, we've done the math, meticulously substituted values, and performed the calculations. Now it's time to step back and interpret our results – what do these findings actually tell us about the given quadratic equation and the numbers 10 and 30? This isn't just about getting a right or wrong answer; it's about understanding the fundamental concept of solutions to quadratic equations.

Here's the lowdown:

  • For x = 10, when we plugged it into (x-27)(x-10)=60, the left side evaluated to 0, while the right side remained 60. Since 0 ≠ 60, we definitively proved that x = 10 is NOT a solution. It simply doesn't satisfy the conditions of the equation. Think of it like trying to fit a square peg in a round hole; it just doesn't work. The equation fails to hold true with 10.
  • For x = 30, on the other hand, when we substituted it into the same equation, the left side evaluated to 60, which perfectly matched the right side, 60. Since 60 = 60, we conclusively determined that x = 30 IS a solution. This value makes the equation true, balancing both sides perfectly. This is exactly what it means to be a solution – the number fulfills the mathematical statement.

This demonstration highlights a critical aspect of algebra: verification. Before you even attempt to solve an equation from scratch, if you're given potential answers, you can always check them. This process not only confirms if a number is a root but also deepens your understanding of how equations work. Every number that is a solution will always make the equation a true statement. Our investigation clearly showed that while 10 might seem like an interesting number in the context of (x-10), the full equation required more, and 30 was the number that delivered the perfect balance. This emphasizes the precision required in mathematics; a near miss is still a miss. Only the exact values that satisfy the equation completely are considered solutions. We’ve now firmly established which of our test values holds true for (x-27)(x-10)=60, and this understanding paves the way for deeper dives into solving these kinds of equations from the ground up.

Beyond Verification: How to Actually Solve Quadratic Equations

Okay, Plastik fam, we've mastered verifying solutions by plugging in numbers, which is a super valuable skill. But what if you weren't given potential answers like 10 and 30? What if you had to find the solutions (or roots) to a quadratic equation like (x-27)(x-10)=60 from scratch? This is where the real fun begins, and thankfully, mathematicians have given us some fantastic tools. Learning how to actually solve quadratic equations is a powerful step in your mathematical journey.

Let's quickly re-express our equation in its standard form: (x-27)(x-10) = 60 First, we expand the left side using the FOIL method (First, Outer, Inner, Last): x x = x² x (-10) = -10x -27 x = -27x -27 (-10) = 270

So, x² - 10x - 27x + 270 = 60 Combine like terms: x² - 37x + 270 = 60

Now, to get it into the standard ax² + bx + c = 0 form, we need to move the 60 from the right side to the left side by subtracting it from both sides: x² - 37x + 270 - 60 = 0 Which simplifies to: x² - 37x + 210 = 0

Now that it's in standard form, we have a few powerhouse methods to find its solutions:

  1. Factoring: This is often the quickest method if it works. You look for two numbers that multiply to c (210) and add up to b (-37). For x² - 37x + 210 = 0, those numbers are -7 and -30. So, the equation factors into: (x - 7)(x - 30) = 0. For this product to be zero, either (x - 7) must be 0 or (x - 30) must be 0. If x - 7 = 0, then x = 7. If x - 30 = 0, then x = 30. Notice something? One of our verified solutions, 30, popped right out! The other solution is 7. This is a super elegant way to solve quadratic equations.

  2. The Quadratic Formula: When factoring isn't straightforward or even possible, the quadratic formula is your best friend. It always works! For ax² + bx + c = 0, the solutions are given by: x = [-b ± √(b² - 4ac)] / 2a For our equation, x² - 37x + 210 = 0, we have a = 1, b = -37, and c = 210. Plugging these values in: x = [ -(-37) ± √((-37)² - 4 * 1 * 210) ] / (2 * 1) x = [ 37 ± √(1369 - 840) ] / 2 x = [ 37 ± √(529) ] / 2 x = [ 37 ± 23 ] / 2 This gives us two solutions: x1 = (37 + 23) / 2 = 60 / 2 = 30 x2 = (37 - 23) / 2 = 14 / 2 = 7 Boom! Again, we found x = 30 and x = 7. The quadratic formula is incredibly powerful for solving quadratic equations of any complexity.

  3. Completing the Square: This method involves transforming the equation so that one side is a perfect square trinomial. While a bit more involved, it’s foundational for understanding the quadratic formula itself and is used in other areas of math.

So, while our initial task was to verify if 10 and 30 were solutions, we've now gone a step further. We've actually solved the equation and found its true solutions: x = 7 and x = 30. This makes perfect sense with our verification results – 30 was a solution, and 10 was not, and now we know why! Understanding these methods provides a complete picture of how to handle quadratic equations, from simply checking values to finding them yourself. Pretty cool, right?

Wrapping Up Our Quadratic Equation Adventure!

And there you have it, Plastik readers! We've journeyed through the world of quadratic equations, tackling the specific challenge of determining whether the values of 10 and 30 are solutions to the equation (x-27)(x-10)=60. What an adventure in solution verification and understanding mathematical roots!

We started by getting a solid grasp of what a quadratic equation is, identifying its x² term, and understanding that its solutions are the values that make the equation true. Then, we meticulously tested x = 10 by substituting it into the equation. Our calculations revealed that 0 ≠ 60, meaning 10 is definitely NOT a solution. It just didn't pass the mathematical sniff test!

Next, we gave x = 30 its turn in the spotlight. After careful substitution and arithmetic, we found that 60 = 60, confirming that 30 IS indeed a solution to our equation. This was a clear demonstration of what it means for a number to satisfy an equation.

Finally, we went above and beyond, exploring how to actually solve quadratic equations from scratch using powerful methods like factoring and the quadratic formula. This allowed us to discover that the true solutions to (x-27)(x-10)=60 are x = 7 and x = 30, perfectly aligning with our verification results. This added layer of understanding really drives home why 10 was rejected and 30 was accepted.

So, the next time you encounter a quadratic equation and are asked to verify potential solutions, you'll know exactly what to do! Remember, guys, math isn't just about memorizing formulas; it's about understanding concepts, applying logical steps, and having the confidence to explore and verify. Keep practicing, keep questioning, and keep those mathematical muscles strong! Thanks for joining us on this enlightening algebraic journey. Until next time, stay curious and keep crunching those numbers!