Solve For X: (x-7)²=36 And (x-1)²=50

Hey guys! Today we're diving into some cool math problems involving quadratic equations. Specifically, we're going to tackle how to find the values of 'x' when you're given equations in the form of $(x-a)^2 = b$. This method is super handy and is often called 'solving by taking the square root'. It's a really fundamental skill in algebra, and once you get the hang of it, you'll see it pop up in all sorts of places. So, let's break down these two examples step-by-step and make sure you're feeling confident about how to solve them. We've got two specific equations to work through: $(x-7)^2=36$ and $(x-1)^2=50$. Both of these follow the same basic structure, which makes learning the process a breeze. We'll start with the first one, $(x-7)^2=36$, and then move on to the second, $(x-1)^2=50$. Get ready to flex those math muscles!

Solving the First Equation: $(x-7)^2=36$

Alright, let's get down to business with our first equation: $(x-7)^2=36$. The goal here is to isolate 'x'. Notice that 'x' is currently part of a squared term, $(x-7)^2$. To undo the squaring, we're going to use the inverse operation, which is taking the square root. When you take the square root of both sides of an equation, you have to remember a very important detail: there are two possible roots – a positive one and a negative one. This is because both a positive number squared and its negative counterpart squared result in the same positive number. So, when we take the square root of 36, we're not just looking for the number that, when multiplied by itself, equals 36 (which is 6), but we also need to consider its negative counterpart (-6).

So, taking the square root of both sides gives us: $\sqrt{(x-7)^2} = \pm\sqrt{36}$. This simplifies to $x-7 = \pm 6$. Now, we've basically split our single equation into two separate, simpler linear equations that we can solve for 'x'.

Case 1: The Positive Root

First, let's consider the positive case: $x-7 = 6$. To find 'x', we just need to add 7 to both sides of the equation. $x = 6 + 7$, which means $x = 13$. So, one possible value for 'x' is 13. Let's quickly check this: $(13-7)^2 = (6)^2 = 36$. Perfect! It works.

Case 2: The Negative Root

Next, we tackle the negative case: $x-7 = -6$. Again, we want to isolate 'x', so we add 7 to both sides. $x = -6 + 7$, which gives us $x = 1$. Let's check this one too: $(1-7)^2 = (-6)^2 = 36$. This also works!

So, for the equation $(x-7)^2=36$, the values of 'x' are 13 and 1. If you were given options, you'd select both A and B for this part. Pretty straightforward, right? The key is remembering that $\pm$ when you take the square root.

Tackling the Second Equation: $(x-1)^2=50$

Now, let's move on to our second equation: $(x-1)^2=50$. We'll use the exact same strategy here. The variable 'x' is inside a squared term, $(x-1)^2$. So, the first step is to take the square root of both sides. Remember our golden rule: we need to account for both the positive and negative roots. This time, however, the square root of 50 isn't a nice, clean whole number like 36 was. That's totally fine! We'll work with the radical.

Taking the square root of both sides gives us: $\sqrt{(x-1)^2} = \pm\sqrt{50}$. This simplifies to $x-1 = \pm\sqrt{50}$. Before we proceed, let's see if we can simplify $\sqrt{50}$. We look for the largest perfect square that divides into 50. That would be 25, because $25 \times 2 = 50$. So, we can rewrite $\sqrt{50}$ as $\sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$. This is a simplified radical form, and it's good practice to always simplify radicals when you can.

So, our equation now is $x-1 = \pm 5\sqrt{2}$. Just like before, this splits into two separate equations.

Case 1: The Positive Root

Let's take the positive case: $x-1 = 5\sqrt{2}$. To solve for 'x', we add 1 to both sides: $x = 1 + 5\sqrt{2}$. This is one of our solutions.

Case 2: The Negative Root

Now for the negative case: $x-1 = -5\sqrt{2}$. Add 1 to both sides to isolate 'x': $x = 1 - 5\sqrt{2}$. This is our second solution.

So, for the equation $(x-1)^2=50$, the values of 'x' are $1 + 5\sqrt{2}$ and $1 - 5\sqrt{2}$. If you were given these options, you would select both A and B. See? Even with radicals involved, the process is the same. You just simplify the radical as much as possible.

Why This Method Works

So, why is this 'taking the square root' method so effective for equations like these? It works because the equation is already set up in a way that makes it easy to isolate the squared term. When you have an expression like $(x-a)^2$, it's essentially a single 'chunk' that's being squared. By taking the square root of both sides, you're directly 'undoing' that squaring operation on the chunk. This leaves you with a much simpler linear equation ($x-a = \pm c$), which is a piece of cake to solve.

Imagine you had to expand $(x-7)^2$ first. You'd get $x^2 - 14x + 49 = 36$. Then you'd have to move the 36 over ($x^2 - 14x + 13 = 0$) and try to factor it or use the quadratic formula. That's way more steps and opens up more chances for errors! The 'taking the square root' method bypasses all that complexity when the equation is in the perfect form $(x-a)^2 = b$. It's an elegant shortcut that highlights the power of understanding inverse operations in algebra. It's like having a secret key to unlock these specific types of problems much faster and more efficiently. So, whenever you see a quadratic equation where the variable term and a constant are grouped together and then squared, and that's it on one side of the equation, think: 'Aha! I can use the square root method!' It’s a game-changer for simplifying these problems and getting to the answer with less hassle. It also helps you appreciate how the structure of an equation can guide you to the most efficient solution path. Keep an eye out for these patterns, guys, they are gold!

Conclusion

There you have it! We've successfully solved two quadratic equations using the square root method. For $(x-7)^2=36$, we found that $x=13$ and $x=1$. And for $(x-1)^2=50$, our solutions are $x=1+5\sqrt{2}$ and $x=1-5\sqrt{2}$. The key takeaways are to always remember the $\pm$ when taking the square root of the constant term, and to simplify any radicals you encounter. This method is super powerful for equations structured like $(x-a)^2 = b$. Keep practicing these, and you'll become a quadratic equation whiz in no time. Math on!