Solving $(\sqrt{9})(8 \sqrt{-28})$: A Step-by-Step Guide

by SLV Team 57 views
Solving $(\sqrt{9})(8 \sqrt{-28})$: A Step-by-Step Guide

Hey guys! Let's dive into this interesting mathematical problem together. We're going to break down how to solve the expression (9)(8βˆ’28)(\sqrt{9})(8 \sqrt{-28}). It looks a little intimidating at first, especially with that negative inside the square root, but trust me, it’s totally manageable. We'll take it step by step, so you can follow along easily and understand each part of the solution. Get ready to sharpen those math skills!

Understanding the Basics

Before we jump right into solving, let’s quickly recap some key concepts. This will make sure we’re all on the same page and the solution makes perfect sense.

Square Roots

First, what exactly is a square root? Simply put, the square root of a number is a value that, when multiplied by itself, gives you the original number. For instance, the square root of 9 is 3 because 3 times 3 equals 9. We write this as 9=3\sqrt{9} = 3. Square roots are the inverse operation of squaring a number. They help us find the base number when we know its square. Understanding square roots is crucial for simplifying expressions and solving equations in algebra and beyond. They pop up everywhere, from geometry to physics, so getting comfortable with them is a great move.

Imaginary Numbers

Now, let's talk about something a bit more intriguing: imaginary numbers. You might be wondering, β€œWhat on earth is an imaginary number?” Well, in the world of real numbers, you can't take the square root of a negative number because no real number multiplied by itself can result in a negative value. That's where imaginary numbers come into play. The basic imaginary unit is denoted by ii, and it's defined as i=βˆ’1i = \sqrt{-1}. This might sound a little abstract, but it opens up a whole new dimension in mathematics!

Using ii, we can express the square root of any negative number. For example, βˆ’28\sqrt{-28} can be rewritten as 28β‹…βˆ’1\sqrt{28} \cdot \sqrt{-1}, which simplifies to 28i\sqrt{28}i. This allows us to work with expressions that would otherwise be undefined in the realm of real numbers. Imaginary numbers are essential in many areas of math and science, including electrical engineering, quantum mechanics, and signal processing. They might seem a bit strange at first, but they're incredibly useful tools!

Simplifying Radicals

Next up, let's tackle simplifying radicals. A radical is just another name for a root, like a square root or a cube root. Simplifying radicals means making the number under the root (the radicand) as small as possible. This usually involves factoring the radicand and pulling out any perfect squares (or perfect cubes, etc., depending on the root). For example, consider 28\sqrt{28}. We can break down 28 into its prime factors: 28=2Γ—2Γ—728 = 2 \times 2 \times 7. Notice that we have a pair of 2s, which means we have a perfect square (22=42^2 = 4) hiding in there. So, we can rewrite 28\sqrt{28} as 4Γ—7\sqrt{4 \times 7}. Then, using the property ab=aβ‹…b\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}, we get 4β‹…7\sqrt{4} \cdot \sqrt{7}, which simplifies to 272\sqrt{7}. This process makes radicals easier to work with and understand.

Breaking Down the Expression (9)(8βˆ’28)(\sqrt{9})(8 \sqrt{-28})

Okay, now that we've refreshed our understanding of square roots, imaginary numbers, and simplifying radicals, let's get back to our original expression: (9)(8βˆ’28)(\sqrt{9})(8 \sqrt{-28}).

Step 1: Simplify 9\sqrt{9}

This is the easiest part! We know that the square root of 9 is 3 because 3Γ—3=93 \times 3 = 9. So, we can replace 9\sqrt{9} with 3 in our expression. Our expression now looks like this: (3)(8βˆ’28)(3)(8 \sqrt{-28}).

Step 2: Deal with the Imaginary Number βˆ’28\sqrt{-28}

Here's where imaginary numbers come into play. We have βˆ’28\sqrt{-28}, which means we're taking the square root of a negative number. As we discussed earlier, this involves imaginary numbers. We can rewrite βˆ’28\sqrt{-28} as 28β‹…βˆ’1\sqrt{28} \cdot \sqrt{-1}. Remember that βˆ’1\sqrt{-1} is defined as ii, the imaginary unit. So, we now have 28i\sqrt{28}i.

Step 3: Simplify 28\sqrt{28}

Now we need to simplify 28\sqrt{28}. To do this, we look for perfect square factors of 28. We can break down 28 into 4Γ—74 \times 7. So, 28\sqrt{28} becomes 4Γ—7\sqrt{4 \times 7}. Using the property ab=aβ‹…b\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}, we can rewrite this as 4β‹…7\sqrt{4} \cdot \sqrt{7}. We know that 4=2\sqrt{4} = 2, so we have 272\sqrt{7}.

Step 4: Substitute Back into the Expression

Let's substitute the simplified form of βˆ’28\sqrt{-28} back into our expression. We found that βˆ’28=27i\sqrt{-28} = 2\sqrt{7}i. So, our expression (3)(8βˆ’28)(3)(8 \sqrt{-28}) becomes (3)(8β‹…27i)(3)(8 \cdot 2\sqrt{7}i).

Putting It All Together

Now that we've simplified all the individual parts, let's combine them to get our final answer.

Step 5: Multiply the Constants

We have (3)(8β‹…27i)(3)(8 \cdot 2\sqrt{7}i). First, let's multiply the constants: 3Γ—8Γ—2=483 \times 8 \times 2 = 48. So, our expression now looks like 487i48\sqrt{7}i.

Step 6: State the Final Answer

That's it! We've simplified the expression as much as possible. The final answer is 487i48\sqrt{7}i.

Why This Matters

You might be thinking,