Question

fan/medal
Multiplying radicals help?

Answer 1

\[(\sqrt{7}+2\sqrt{5})(3\sqrt{7}-6\sqrt{5})\]

Answer 2

Well, just distribute them..

\(\sqrt7 \times 3 \sqrt 7\)
\(\sqrt7 \times -6 \sqrt 5\)

\(2 \sqrt 5 \times 3 \sqrt 7\)
\(2 \sqrt 5 \times -6 \sqrt 5\)

Answer 3

I know how to distribute but I don't know how to multiply the radicals

Answer 4

\(\sqrt 7 \times 3 \sqrt 7\)

Multiply the square roots together:

\(3 \sqrt {49}\)

49 is a perfect square, so it can be squared.

The square root of 49 is 7..

\((3)(7)\)

\(21\)

Answer 5

Ok great. so you can multiply them is the square roots aren't the same?

Answer 6

Now can you do that for \(-6 \sqrt 5\)?

Answer 7

\[-6\sqrt{35}\]

Answer 8

Well, you multiply the numbers in the square roots together.

Like for:

\(2\sqrt 7 \times 3 \sqrt 2\)

We multiply the whole numbers together and the square roots, then we combine them:

\(2 \times 3 = 6\)
\(7 \times 2 = 14\)

Therefore we have:

\(6 \sqrt {14}\)

Answer 9

Yep, \(-6 \sqrt {35}\) is correct.

Now can you multiply this one?

\(2 \sqrt 5 \times 3 \sqrt 7\)

Answer 10

@ariabstr

Answer 11

\[6\sqrt{35}\]

Answer 12

Yep.

Answer 13

Now for the last one:

\(2 \sqrt 5 \times -6 \sqrt 5\)

Answer 14

\[-12\sqrt{25}\]

Answer 15

Yep, and 25 is a perfect square..so you can simplify it even more.

Answer 16

so then you have \[21-6\sqrt{35}+6\sqrt{35}-12\sqrt{25}\]

Answer 17

\[21-12\sqrt{25}\]

Answer 18

Yes..but we can find the square root of 25..which is 5.

\(-12 \sqrt {25}\)

\((-12)(5)\)

\(-60\)

Answer 19

-39?

Answer 20

So we have:

\(21 -6 \sqrt{35} + 6 \sqrt {35} - 60\)

\(-6 \sqrt{35}\) and \(6 \sqrt {35}\) cancel out..leaving you with:

\(21 - 60\)

\(-39~\Large\color{lime} \checkmark\)

Answer 21

Ok I understand now. Thank you so much

Answer 22

No problem! :D

Answer 23

Could you help with one thats similar?

Answer 24

\[\left( 135x ^{12}y ^{10} \right)^{\left(\begin{matrix}1 \\ 3\end{matrix}\right)}\]

Answer 25

thats supposed to be 1/3