Solve Quadratic Equations: $15x^2 - 56 = 88 + 6x^2$

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of solving quadratic equations. You know, those equations that look a bit like a rollercoaster with their squared terms? Well, we've got a doozy for you: $15 x^2-56=88+6 x^2$. Our mission, should we choose to accept it (and we totally do!), is to find all the values of $x$ that make this equation true. We'll be checking the options provided: A. $x=4$, B. $x=-4$, C. $x=12$, D. $x=-12$, E. $x=-\sqrt{2}$, and F. $x=\sqrt{2}$. Get ready to flex those math muscles, because we're about to break it down step-by-step. It's going to be a wild ride, but don't worry, we'll keep it light and fun. Remember, understanding how to solve these types of equations is super useful, not just for acing your exams but also for tackling real-world problems that involve curves, projections, or anything with a quadratic relationship. So, grab your favorite drink, get comfy, and let's unravel this mathematical mystery together!

Understanding the Quadratic Equation

Alright, let's get down to business with our equation: $15 x^2-56=88+6 x^2$. The first thing we wanna do when we see an equation like this is to simplify it. Our goal is to get all the $x^2$ terms on one side and all the constant numbers on the other side. Think of it like organizing your closet – everything has its place! So, let's start by moving the $6x^2$ term from the right side to the left side. To do this, we subtract $6x^2$ from both sides of the equation. This gives us: $(15 x^2 - 6x^2) - 56 = 88 + (6x^2 - 6x^2)$. Simplifying this, we get $9x^2 - 56 = 88$. See? Already looking a lot cleaner, right? Now, we need to get that $-56$ away from the $9x^2$ term. We do this by adding $56$ to both sides of the equation. So, $9x^2 - 56 + 56 = 88 + 56$. This leaves us with $9x^2 = 144$. Boom! We're one step closer to finding our mysterious $x$ values. This simplified form, $9x^2 = 144$, is where the real magic begins. It's a pure quadratic equation, meaning it only has an $x^2$ term and a constant. This makes it way easier to solve than a full quadratic equation with an $x$ term too. We're essentially isolating the $x^2$ term to figure out what $x^2$ equals, and then we'll take the square root to find $x$. It's a process that requires a bit of algebraic manipulation, but nothing we can't handle, right? Keep that concentration up, because the next steps are crucial in uncovering the solutions we're looking for.

Isolating $x^2$ and Finding the Solutions

Now that we've got our equation down to $9x^2 = 144$, the next logical step is to isolate $x^2$. To do this, we need to get rid of that $9$ that's multiplying $x^2$. We achieve this by dividing both sides of the equation by $9$. So, we have rac{9x^2}{9} = rac{144}{9}. Performing the division, we find that $x^2 = 16$. This is a super important step, guys! We've figured out that $x$ squared equals $16$. Now, to find the actual value of $x$, we need to take the square root of both sides. Remember, when you take the square root of a number, there are two possible answers: a positive one and a negative one. This is because both a positive number squared and a negative number squared result in a positive number. So, when we take the square root of $16$, we get $x = \pm\sqrt{16}$. The square root of $16$ is $4$. Therefore, our solutions for $x$ are $x = 4$ and $x = -4$. We've officially found the values of $x$ that satisfy the original equation! It's always crucial to remember that ± sign when taking the square root in equations like this. Forgetting it can lead to missing one of the valid solutions. We've successfully navigated through the algebraic steps, combined like terms, and used the power of the square root to unlock the values of $x$. It’s a testament to how systematic algebra works – each step builds upon the last, leading us closer to the solution. This process solidifies the understanding that quadratic equations, especially those in this simplified form, often yield two distinct answers, reflecting the parabolic nature of their graphs.

Checking the Options and Verifying Our Answers

Okay, we've done the heavy lifting and found our potential solutions: $x=4$ and $x=-4$. Now, let's double-check our work by plugging these values back into the original equation: $15 x^2-56=88+6 x^2$. This verification step is super important, like proofreading an essay, to make sure everything is accurate. Let's start with $x=4$. Plugging it in, we get: $15(4)^2 - 56 = 88 + 6(4)^2$. Calculate the squares: $4^2 = 16$. So, the equation becomes $15(16) - 56 = 88 + 6(16)$. Now, perform the multiplications: $15 imes 16 = 240$ and $6 imes 16 = 96$. So, we have $240 - 56 = 88 + 96$. Finally, perform the subtractions and additions: $240 - 56 = 184$ and $88 + 96 = 184$. Since $184 = 184$, our solution $x=4$ is correct!

Now, let's check $x=-4$. Plugging it in: $15(-4)^2 - 56 = 88 + 6(-4)^2$. Remember, squaring a negative number makes it positive, so $(-4)^2 = 16$. The equation becomes $15(16) - 56 = 88 + 6(16)$. This is exactly the same calculation as for $x=4$! So, we get $240 - 56 = 88 + 96$, which simplifies to $184 = 184$. Therefore, our solution $x=-4$ is also correct!

Let's quickly look at the other options to see why they don't work. If we tried $x=12$ or $x=-12$, their squares would be $144$. Plugging $x^2=144$ into $9x^2=144$ gives $9(144)=144$, which is false. If we tried $x=\sqrt{2}$ or $x=-\sqrt{2}$, their squares would be $2$. Plugging $x^2=2$ into $9x^2=144$ gives $9(2)=144$, which is $18=144$, also false. This verification process is crucial, not just for confirming our answers but also for understanding why other values are not solutions. It reinforces the algebraic manipulations we performed and gives us confidence in our final results. It’s a satisfying feeling when your calculated solutions hold true when plugged back into the original equation, confirming the accuracy of your mathematical journey.

Conclusion: The Solutions to the Equation

So, after all that hard work and careful checking, we've definitively found the values of $x$ that make the equation $15 x^2-56=88+6 x^2$ true. We simplified the equation, isolated the $x^2$ term, and solved for $x$, remembering to account for both positive and negative roots. Our verification process confirmed that $x=4$ and $x=-4$ are indeed the correct solutions. Therefore, the options that apply are A. $x=4$ and B. $x=-4$. It’s awesome when math problems have clear-cut answers, and this one definitely does! Remember, tackling quadratic equations is a fundamental skill in algebra, and with practice, you'll become a pro at simplifying, solving, and verifying. Keep practicing these concepts, guys, and don't hesitate to review the steps if you get stuck. Understanding the principles behind solving these equations not only helps in academic settings but also builds a strong foundation for more complex mathematical and scientific applications. We hope this breakdown was helpful and kept you engaged. Until next time, keep those brains buzzing and keep exploring the fascinating world of mathematics with Plastik Magazine!