Matching Equations: Solve For 'x' With Given Values

Hey guys! Today, we're diving into the exciting world of equation-solving. Our mission, should we choose to accept it, is to match each equation with the correct value of x that makes it true. We've got a set of values to choose from: 1, 5, 2, 18, 9, and -3. And we have three intriguing equations that are just begging to be solved. Buckle up, because we're about to embark on a mathematical adventure!

Equation 1: $\sqrt{x^2+7}=4$

Okay, let's kick things off with our first equation: $\sqrt{x^2+7}=4$. This equation involves a square root, which might seem intimidating at first, but fear not! We can tackle this step by step. The key here is to remember how to get rid of that pesky square root. What's the inverse operation of taking a square root? You guessed it – squaring!

So, our first move is to square both sides of the equation. This will eliminate the square root on the left side and give us a more manageable equation to work with. Squaring both sides, we get:

$(\sqrt{x^2+7})^2 = 4^2$

This simplifies to:

$x^2 + 7 = 16$

Now we're cooking! We've transformed our original equation into a much simpler quadratic equation. To solve for x, we need to isolate the $x^2$ term. We can do this by subtracting 7 from both sides of the equation:

$x^2 + 7 - 7 = 16 - 7$

Which gives us:

$x^2 = 9$

Almost there! We now have $x^2$ equal to 9. To find x, we need to take the square root of both sides. But here's a crucial point to remember: when we take the square root of both sides of an equation, we need to consider both the positive and negative roots. This is because both 3 squared and -3 squared equal 9.

So, taking the square root of both sides, we get:

$x = \pm \sqrt{9}$

Which means:

$x = \pm 3$

This gives us two potential solutions: x = 3 and x = -3. But wait! We need to check these solutions to make sure they actually work in the original equation. Sometimes, when dealing with square roots, we can get extraneous solutions – solutions that satisfy the transformed equation but not the original one. So, let's plug each of these values back into the original equation, $\sqrt{x^2+7}=4$, and see what happens.

Checking x = 3:

$\sqrt{(3)^2 + 7} = \sqrt{9 + 7} = \sqrt{16} = 4$

Yep, x = 3 works!

Checking x = -3:

$\sqrt{(-3)^2 + 7} = \sqrt{9 + 7} = \sqrt{16} = 4$

And x = -3 works too!

So, we have two solutions for this equation: x = 3 and x = -3. Looking at our list of possible values, we see that -3 is one of the options. Therefore, the value of x that satisfies the first equation is -3. Pat yourself on the back – you've solved the first equation!

Equation 2: $\sqrt[3]{1-x}=-1$

Alright, let's move on to the second equation: $\sqrt[3]{1-x}=-1$. This time, we're dealing with a cube root instead of a square root. But don't fret, the same principle applies! We need to figure out how to undo the cube root. What operation is the inverse of taking a cube root? You guessed it again – cubing!

To eliminate the cube root, we'll cube both sides of the equation. This will give us a simpler equation to solve. Cubing both sides, we get:

$(\sqrt[3]{1-x})^3 = (-1)^3$

This simplifies to:

$1 - x = -1$

Now we have a linear equation, which is much easier to handle. Our goal is to isolate x. We can start by subtracting 1 from both sides:

$1 - x - 1 = -1 - 1$

This gives us:

$-x = -2$

Almost there! To get x by itself, we can multiply both sides by -1:

$(-1)(-x) = (-1)(-2)$

Which results in:

$x = 2$

We've found a potential solution: x = 2. Now, let's check if it actually works in the original equation, $\sqrt[3]{1-x}=-1$.

Checking x = 2:

$\sqrt[3]{1-2} = \sqrt[3]{-1} = -1$

Great! x = 2 satisfies the equation. Looking at our list of possible values, we see that 2 is indeed one of the options. So, the value of x that satisfies the second equation is 2. Another equation down – you're on a roll!

Equation 3: $(x-2)^{\frac{1}{4}}=2$

Let's tackle the final equation: $(x-2)^{\frac{1}{4}}=2$. This equation might look a bit different, but it's actually quite similar to the previous ones. Remember that a fractional exponent like $\frac{1}{4}$ represents a root. In this case, $(x-2)^{\frac{1}{4}}$ is the same as the fourth root of (x-2), or $\sqrt[4]{x-2}$.

So, our equation is essentially $\sqrt[4]{x-2} = 2$. To get rid of the fourth root, we need to raise both sides of the equation to the power of 4. Why the power of 4? Because that's the inverse operation of taking the fourth root!

Raising both sides to the power of 4, we get:

$((x-2)^{\frac{1}{4}})^4 = 2^4$

This simplifies to:

$x - 2 = 16$

Now we have a simple linear equation. To solve for x, we just need to add 2 to both sides:

$x - 2 + 2 = 16 + 2$

This gives us:

$x = 18$

We've found a potential solution: x = 18. Let's check if it works in the original equation, $(x-2)^{\frac{1}{4}}=2$.

Checking x = 18:

$(18 - 2)^{\frac{1}{4}} = (16)^{\frac{1}{4}} = \sqrt[4]{16} = 2$

Perfect! x = 18 satisfies the equation. And guess what? 18 is one of the values in our list! So, the value of x that satisfies the third equation is 18.

Conclusion

Fantastic job, guys! We've successfully matched each equation with the value of x that satisfies it. To recap, here's what we found:

• $\sqrt{x^2+7}=4$ is satisfied by x = -3.
• $\sqrt[3]{1-x}=-1$ is satisfied by x = 2.
• $(x-2)^{\frac{1}{4}}=2$ is satisfied by x = 18.

By carefully applying inverse operations and checking our solutions, we were able to conquer these equations. Keep practicing, and you'll become a master equation solver in no time! Remember, the key is to break down complex problems into smaller, manageable steps. You got this!