Multiplying Radicals of the Same Root - Problem 1


Distributing in a square root is exactly the same as distributing in any other number okay? Square root of 3 in this case is a number. It's not a good number you need to plug in the calculated figure at exactly what it is, but it's still just a number and it's on right.

So I think if we have like 2 times 3 plus x, we know that this 2 has to get distributed into both things, this goes into both, we end up with 6 plus 2x, that's easy enough. With a square root it's exactly the same idea. The square root on the outside is going to have to get distributed into both square roots on the inside.

So let's carry to this little example here and we just distribute this in, square root of 3 times the square root of 5, we have the same square roots, so we can combine it turns into square root of 15. Square root of 3 times the square root of 3, any time you're multiplying two square roots and they're the same thing, we just cancel the square roots and we end up with 3. Square root of 15 can't be simplified at all, there's no perfect squares that go into it, so by distributing in the term on the outside we were able to simplify this expression.

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