SOLUTION: sqrt98 times 2sqrt 48

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Question 388444: sqrt98 times 2sqrt 48
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%2898%29%2A2sqrt%2848%29
Square roots are a little like fractions in that when you multiply them it is to your advantage to simplify before you actually multiply. Simplifying is easier before the multiplication than it will be after the multiplication.

So before we multiply we will simplify the square roots. Simplifying square roots starts with finding perfect square factors (other than 1) of the radicands (the expression within a radical is called a radicand). Each of these radciands has a perfect square factor:
sqrt%2849%2A2%29%2A2sqrt%2816%2A3%29
Now we can use a property of radicals, root%28a%2C+p%2Aq%29+=+root%28a%2C+p%29%2Aroot%28a%2C+q%29, to separate the perfect square factors into their own square roots:
sqrt%2849%29%2Asqrt%282%29%2A2sqrt%2816%29%2Asqrt%283%29
The square roots of the perfect squares can be simplified:
7%2Asqrt%282%29%2A2%2A4%2Asqrt%283%29
This is all multiplication. And since multiplication is Commutative we can multiply these numbers in any order. We can multiply the integers and, using the same property of radicals above only in reverse, we can multiply the square roots giving us:
56%2Asqrt%282%2A3%29
which simplifies to
56%2Asqrt%286%29
Since the square root of 6 has no perfect square factors (other than 1), it will not simplify further.

If we had not simplified before we multiplied we would get
2sqrt%284704%29
after multiplying. But this is not the simplified answer. We have to look for perfect square factors of 4704!? If done correctly you would still end up with:
56%2Asqrt%286%29
I hope you agree that simplifying first made the problem easier.