Math / Algebra

QuestionSquare root addition or subtraction with three terms
Simplify. $2 \sqrt{27}+15 \sqrt{3}-\sqrt{12}$

Studdy Solution

STEP 1

What is this asking? We've got a bunch of square roots here, and we need to simplify them by adding and subtracting them together! Watch out! We can only add or subtract square roots if the stuff inside the square root is the same.
So we'll need to simplify first!

STEP 2

1. Simplify the square roots
2. Combine like terms

STEP 3

Alright, let's break down that first term, $2 \sqrt{27}$ .
We're looking for perfect squares hiding inside 27.
Think: what perfect square goes into 27?
It's 9!
We can rewrite 27 as $9 \cdot 3$ .

STEP 4

So, $2 \sqrt{27}$ becomes $2 \sqrt{9 \cdot 3}$ .
Since $\sqrt{9}$ is 3, we can pull that 3 out front and multiply it by the 2 that's already there.
This gives us $2 \cdot 3 \sqrt{3}$ , which simplifies to $6 \sqrt{3}$ .
Awesome!

STEP 5

The second term, $15 \sqrt{3}$ , is already simplified.
The number inside the square root, 3, is a prime number, so there are no perfect square factors to pull out.
We can leave it as is.

STEP 6

Now, let's tackle the last term, $\sqrt{12}$ .
We need to find a perfect square inside 12.
4 goes into 12!
We can rewrite 12 as $4 \cdot 3$ .

STEP 7

So, $\sqrt{12}$ becomes $\sqrt{4 \cdot 3}$ .
Since $\sqrt{4}$ is 2, we can pull that 2 out front.
This gives us $2 \sqrt{3}$ .

STEP 8

Now, we've got $6 \sqrt{3} + 15 \sqrt{3} - 2 \sqrt{3}$ .
Since all the square roots have the same number inside (that's the 3), we can add and subtract the numbers outside the square roots!

STEP 9

Think of it like this: we have 6 "things" plus 15 "things" minus 2 "things".
Where "thing" is $\sqrt{3}$ .
So, $6 + 15 - 2$ gives us 19.

STEP 10

Therefore, our final simplified expression is $19 \sqrt{3}$ !

STEP 11

$19 \sqrt{3}$

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QuestionSquare root addition or subtraction with three terms
Simplify. $2 \sqrt{27}+15 \sqrt{3}-\sqrt{12}$

Studdy Solution

STEP 1

What is this asking? We've got a bunch of square roots here, and we need to simplify them by adding and subtracting them together! Watch out! We can only add or subtract square roots if the stuff inside the square root is the same.
So we'll need to simplify first!

STEP 2

1. Simplify the square roots
2. Combine like terms

STEP 3

Alright, let's break down that first term, $2 \sqrt{27}$ .
We're looking for perfect squares hiding inside 27.
Think: what perfect square goes into 27?
It's 9!
We can rewrite 27 as $9 \cdot 3$ .

STEP 4

So, $2 \sqrt{27}$ becomes $2 \sqrt{9 \cdot 3}$ .
Since $\sqrt{9}$ is 3, we can pull that 3 out front and multiply it by the 2 that's already there.
This gives us $2 \cdot 3 \sqrt{3}$ , which simplifies to $6 \sqrt{3}$ .
Awesome!

STEP 5

The second term, $15 \sqrt{3}$ , is already simplified.
The number inside the square root, 3, is a prime number, so there are no perfect square factors to pull out.
We can leave it as is.

STEP 6

Now, let's tackle the last term, $\sqrt{12}$ .
We need to find a perfect square inside 12.
4 goes into 12!
We can rewrite 12 as $4 \cdot 3$ .

STEP 7

So, $\sqrt{12}$ becomes $\sqrt{4 \cdot 3}$ .
Since $\sqrt{4}$ is 2, we can pull that 2 out front.
This gives us $2 \sqrt{3}$ .

STEP 8

Now, we've got $6 \sqrt{3} + 15 \sqrt{3} - 2 \sqrt{3}$ .
Since all the square roots have the same number inside (that's the 3), we can add and subtract the numbers outside the square roots!

STEP 9

Think of it like this: we have 6 "things" plus 15 "things" minus 2 "things".
Where "thing" is $\sqrt{3}$ .
So, $6 + 15 - 2$ gives us 19.

STEP 10

Therefore, our final simplified expression is $19 \sqrt{3}$ !

STEP 11

$19 \sqrt{3}$

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QuestionSquare root addition or subtraction with three termsSimplify. 227+153−122 \sqrt{27}+15 \sqrt{3}-\sqrt{12}227​+153​−12​

Studdy Solution

STEP 1

What is this asking? We've got a bunch of square roots here, and we need to simplify them by adding and subtracting them together! Watch out! We can only add or subtract square roots if the stuff *inside* the square root is the same.So we'll need to simplify first!

STEP 2

1. Simplify the square roots2. Combine like terms

STEP 3

Alright, let's **break down** that first term, 2272 \sqrt{27}227​.We're looking for **perfect squares** hiding inside 27.Think: what perfect square goes into 27?It's 9!We can rewrite 27 as 9⋅39 \cdot 39⋅3.

STEP 4

So, 2272 \sqrt{27}227​ becomes 29⋅32 \sqrt{9 \cdot 3}29⋅3​.Since 9\sqrt{9}9​ is 3, we can pull that **3 out front** and multiply it by the 2 that's already there.This gives us 2⋅332 \cdot 3 \sqrt{3}2⋅33​, which simplifies to 636 \sqrt{3}63​.Awesome!

STEP 5

The second term, 15315 \sqrt{3}153​, is already simplified.The number inside the square root, 3, is a **prime number**, so there are no perfect square factors to pull out.We can leave it as is.

STEP 6

Now, let's tackle the last term, 12\sqrt{12}12​.We need to find a **perfect square** inside 12.4 goes into 12!We can rewrite 12 as 4⋅34 \cdot 34⋅3.

STEP 7

So, 12\sqrt{12}12​ becomes 4⋅3\sqrt{4 \cdot 3}4⋅3​.Since 4\sqrt{4}4​ is 2, we can pull that **2 out front**.This gives us 232 \sqrt{3}23​.

STEP 8

Now, we've got 63+153−236 \sqrt{3} + 15 \sqrt{3} - 2 \sqrt{3}63​+153​−23​.Since all the square roots have the same number inside (that's the **3**), we can add and subtract the numbers *outside* the square roots!

STEP 9

Think of it like this: we have 6 "things" plus 15 "things" minus 2 "things".Where "thing" is 3\sqrt{3}3​.So, 6+15−26 + 15 - 26+15−2 gives us **19**.

STEP 10

Therefore, our **final simplified expression** is 19319 \sqrt{3}193​!

STEP 11

19319 \sqrt{3}193​

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QuestionSquare root addition or subtraction with three terms
Simplify. $2 \sqrt{27}+15 \sqrt{3}-\sqrt{12}$

What is this asking? We've got a bunch of square roots here, and we need to simplify them by adding and subtracting them together! Watch out! We can only add or subtract square roots if the stuff inside the square root is the same.
So we'll need to simplify first!

1. Simplify the square roots
2. Combine like terms

Alright, let's break down that first term, $2 \sqrt{27}$ .
We're looking for perfect squares hiding inside 27.
Think: what perfect square goes into 27?
It's 9!
We can rewrite 27 as $9 \cdot 3$ .

So, $2 \sqrt{27}$ becomes $2 \sqrt{9 \cdot 3}$ .
Since $\sqrt{9}$ is 3, we can pull that 3 out front and multiply it by the 2 that's already there.
This gives us $2 \cdot 3 \sqrt{3}$ , which simplifies to $6 \sqrt{3}$ .
Awesome!

The second term, $15 \sqrt{3}$ , is already simplified.
The number inside the square root, 3, is a prime number, so there are no perfect square factors to pull out.
We can leave it as is.

Now, let's tackle the last term, $\sqrt{12}$ .
We need to find a perfect square inside 12.
4 goes into 12!
We can rewrite 12 as $4 \cdot 3$ .

So, $\sqrt{12}$ becomes $\sqrt{4 \cdot 3}$ .
Since $\sqrt{4}$ is 2, we can pull that 2 out front.
This gives us $2 \sqrt{3}$ .

Now, we've got $6 \sqrt{3} + 15 \sqrt{3} - 2 \sqrt{3}$ .
Since all the square roots have the same number inside (that's the 3), we can add and subtract the numbers outside the square roots!

Think of it like this: we have 6 "things" plus 15 "things" minus 2 "things".
Where "thing" is $\sqrt{3}$ .
So, $6 + 15 - 2$ gives us 19.

Therefore, our final simplified expression is $19 \sqrt{3}$ !

$19 \sqrt{3}$