I can't figure these out and any help would be appreciated! Thank you!

1. 9(√6-2)

2. (6√2+8√3)(√2+9√3)

3. (√6+x)^{2}

4. (^{3}√4-6)(^{3}√2+6)

5. (√x-3+5)^{2}

6. 7y√20y+9√20y^{3}

I can't figure these out and any help would be appreciated! Thank you!

1. 9(√6-2)

2. (6√2+8√3)(√2+9√3)

3. (√6+x)^{2}

4. (^{3}√4-6)(^{3}√2+6)

5. (√x-3+5)^{2}

6. 7y√20y+9√20y^{3}

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Hi Courtnee,

I am not going to provide you with the answers to all of these problems because it's your homework not mine :), but I will help you learn how to solve problems such as this, so I will illustrate with #2. It will seem as if I am over using parentheses, but because of the limitations of this forum I will need to use them to give clarity and ensure accuracy.

(6√(2)+8√(3))(√(2)+9√(3))

This problem is both multiplication of square roots as well as addition...so let's discuss what the rules are first.

1. You can only add or subtract square roots when the number under the radical (**√()**) is the same e.g.

**√(2) + √(2) = 2√(2). **

This reads the square root of 2 plus the square root of 2 equals 2 times the square root of 2...or 2 copies of the square root of 2,
** like 2x is x + x**.

Subtraction would be the same way...

**2√(2) - √(2) = √(2)**

You can't add things that aren't alike...e.g. the same # is not under the radical, so...

**√(2) + 9√(3)** = * √(2) + 9√(3) * ==> This has to stay as-is because √(2) & √(3) are not the same, they are apples and oranges...

2. When you multiply and divide square roots the rules change a bit...

==> You can multiply only the numbers that are under the radical (**√()**) e.g.

√(2) * √(3) = √(6)

...but an easier way to realize this is √(2) * √(2) = √(4) which equals 2...but we know that.

Try that one on your calculator just to check, by finding the square root of 2 and then square it. You may get 3.99999 something, but that's because the calculator rounded the square root of 2.

3. Division works the same way as multiplication, except you may end up with a fraction that has to stay.

**√(4) ÷ √(2) is really √(4/2) = √(2)**...remember we can write the √(4) as √(2)*√(2)...that makes cancellation easy

...but what if it was the **√(2) ÷ √(4)**?

Then the answer would be

**√(2)**

** 2**

Even still what if it was worse...**√(8) ÷ √(6)**...I know gross!

But use the same principles and you get...

**2√(2) **

** √(6)**

Remember 4*2 = 8 and √(4) is 2, so we can take that out of the radical.

If we factor √(6) we get √(2) * √(3)...well just like we did before we can do cancellations

**2 * √(2) = 2 * √(2) = 2
**

**√(3) * √(2) = √(3) * √(2) = √(3)**

That answer is correct, but we generally don't like square roots or roots of any kind in the denominator...so we multiply by the same thing that is in the denominator so we can get an answer that looks like this:

** 2 * √(3) =
2 √(3) **

**√(3) * √(3) = 3**

Now getting back to the problem...you can use matrix multiplcation (box method), the concepts from FOIL, or just remember that each term on has to be multiplied by the other terms based on the
* distributive property*...

**(6√(2))* (√(2) + 9√(3))** and **(8√(3))*(√(2) + 9√(3))**.

**Add up the LIKE TERMS and VOILA you have your answer!**

Richard B.

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## Comments

Hi Courtnee,

For #1 can you verify that you mean is 9 times the quantity (square root of 6) - 2. I put the parentheses in because I don't want to misinterpret. The same goes for the others except #2.