Simplify $\sqrt{3 C^7} \cdot \sqrt{75 C}$ For $c \geq 0$

Hey math whizzes and number crunchers! Today, we're diving into a cool problem that involves simplifying radicals. We're going to tackle the expression $\\sqrt{3 c^7} \cdot \sqrt{75 c}$, and the key condition here is that $c \geq 0$. This little detail is super important because it means we don't have to worry about dealing with imaginary numbers when we take the square root of variables. So, grab your calculators (or just your brilliant brains!), and let's break this down step-by-step.

Understanding Radicals and Simplification

Before we jump into the thick of it, let's quickly chat about what simplifying radicals even means. When we talk about simplifying a radical expression, like our square root problem, we're basically trying to make it as neat and tidy as possible. This usually involves a few things:

1. Extracting Perfect Squares: If there are any perfect square factors inside the radical (the number under the $\sqrt{}$ sign), we want to pull them out. For example, $\sqrt{16}$ simplifies to 4 because 16 is $4^2$. Similarly, $\sqrt{x^4}$ simplifies to $x^2$ because $x^4 = (x^2)^2$.
2. Combining Terms: If we have multiple radicals multiplied together, we can often combine them under a single radical sign using the property $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$.
3. Rationalizing the Denominator: (Not applicable here, but good to know!) If there's a radical in the denominator, we usually try to get rid of it.

Our problem involves the multiplication of two square roots, so we'll definitely be using that second property. The fact that $c \geq 0$ simplifies things immensely. When you have a variable raised to an odd power inside a square root, like $c^7$, you can rewrite it as $c^6 \cdot c$. Then, $\sqrt{c^7}$ becomes $\sqrt{c^6 \cdot c} = \sqrt{c^6} \cdot \sqrt{c} = c^3 \sqrt{c}$. The $c^6$ part is a perfect square because 6 is an even exponent. If $c$ could be negative, we'd have to be more careful with absolute values, but since $c \geq 0$, everything stays straightforward.

Step 1: Combine the Radicals

The first move we're going to make is to combine the two square roots into one. Remember that property we talked about? $\\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$. We're going to apply this to our expression:

$$\\sqrt{3 c^7} \cdot \sqrt{75 c} = \sqrt{(3 c^7) \cdot (75 c)}$$

Now, we just need to multiply the stuff inside the radical. Let's multiply the numbers and the variables separately:

• Numbers: $3 \cdot 75$. Hmm, what's that? $3 \times 70 = 210$, and $3 \times 5 = 15$. So, $210 + 15 = 225$. Easy peasy!
• Variables: $c^7 \cdot c$. When you multiply terms with the same base, you add their exponents. Remember that $c$ is the same as $c^1$. So, $c^7 \cdot c^1 = c^{7+1} = c^8$.

Putting it all together, the expression inside the radical becomes $225 c^8$. So, our combined radical is:

$$\\sqrt{225 c^8}$$

See? That looks a lot simpler already! We've gone from two separate radicals to just one. High five!

Step 2: Simplify the Combined Radical

Now that we have $\\sqrt{225 c^8}$, our next job is to simplify this beast. We need to find the square root of $225$ and the square root of $c^8$. Let's tackle them one by one.

• Simplifying $\\sqrt{225}$: Do you recognize 225? If you're familiar with perfect squares, you might know that $15^2 = 225$. So, $\sqrt{225} = 15$. If you didn't know that off the top of your head, you could use prime factorization to find it. The prime factors of 225 are $3 \times 3 \times 5 \times 5$, which is $(3^2) \times (5^2)$. When you take the square root, you get $3 \times 5 = 15$. Bingo!

• Simplifying $\\sqrt{c^8}$: This one is even more straightforward, especially since we know $c \geq 0$. Remember the rule for square roots of variables with even exponents: $\sqrt{x^{2n}} = x^n$. In our case, the exponent is 8, which is $2 \times 4$. So, $\sqrt{c^8} = c^{8/2} = c^4$. Since $c \geq 0$, we don't need to worry about absolute value bars.

Now, we just put these simplified parts back together. The square root of the product is the product of the square roots:

$$\\sqrt{225 c^8} = \sqrt{225} \cdot \sqrt{c^8} = 15 \cdot c^4$$

And there you have it! The simplified form of the expression is $15 c^4$. How cool is that? We took a somewhat complex-looking expression and turned it into something super clean and easy to work with. This is why math is awesome, guys!

Alternative Approach: Simplify First, Then Multiply

Sometimes, it's helpful to simplify each radical before you combine them. Let's see if that gives us the same answer. This approach is particularly useful when the exponents inside the radicals are odd, like $c^7$ in our first term.

Let's look at the first radical: $\sqrt{3 c^7}$.

• We can rewrite $c^7$ as $c^6 \cdot c$. Since $c^6 = (c^3)^2$, it's a perfect square.
• So, $\sqrt{3 c^7} = \sqrt{3 \cdot c^6 \cdot c}$.
• Using the radical property $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$, we can separate this:

  $$\\sqrt{3 \cdot c^6 \cdot c} = \sqrt{3} \cdot \sqrt{c^6} \cdot \sqrt{c}$$

• We know $\sqrt{c^6} = c^{6/2} = c^3$. (Again, since $c \geq 0$, no absolute value needed).
• So, the first radical simplifies to: $\sqrt{3} \cdot c^3 \cdot \sqrt{c}$, or $c^3 \sqrt{3c}$.

Now let's look at the second radical: $\sqrt{75 c}$.

• We need to find perfect square factors of 75. Let's break 75 down: $75 = 3 \times 25$. And 25 is a perfect square ($5^2$)!
• So, $\sqrt{75 c} = \sqrt{25 \cdot 3 \cdot c}$.
• Separating this using our radical property:

  $$\\sqrt{25 \cdot 3 \cdot c} = \sqrt{25} \cdot \sqrt{3c}$$

• We know $\sqrt{25} = 5$.
• So, the second radical simplifies to: $5 \sqrt{3c}$.

Alright, so now we have our two simplified radicals: $c^3 \sqrt{3c}$ and $5 \sqrt{3c}$. Let's multiply them together:

$$(c^3 \sqrt{3c}) \cdot (5 \sqrt{3c})$$

To multiply these, we multiply the parts outside the radicals together and the parts inside the radicals together:

• Outside: $c^3 \cdot 5 = 5c^3$.
• Inside: $\sqrt{3c} \cdot \sqrt{3c}$. When you multiply a radical by itself, you just get the number inside the radical. So, $\sqrt{3c} \cdot \sqrt{3c} = 3c$.

Putting it all together:

$$(5c^3) \cdot (3c) = 15 c^{3+1} = 15 c^4$$

Voila! We got the exact same answer, $15 c^4$, using the