SOLUTION: I have had a hard time figuring out this problem. Please help. The problem is the square root of 3 divided by the square root of 3 + the square root of 2. I think that the squa

Algebra ->  Square-cubic-other-roots -> SOLUTION: I have had a hard time figuring out this problem. Please help. The problem is the square root of 3 divided by the square root of 3 + the square root of 2. I think that the squa      Log On


   



Question 48331: I have had a hard time figuring out this problem. Please help. The problem is the square root of 3 divided by the square root of 3 + the square root of 2. I think that the square roots of t3 can cancel out and the answer be 1/square root of two.
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
DEFINITELY DEFINITELY NOT!!!!!!!!

+%28sqrt%283%29%29%2F%28sqrt%283%29+%2B+sqrt%282%29%29+ is treated just like +%283%29%2F%283%2B2%29+ which would be 3/5. You would not say cancel the 3's leaving 1/2.

Treat it as any normal fraction and apply Rules of Fractions and not your own made up rules :-)

Now, with radicals (or surds) ie square roots, we tend to want no square roots on the denominator. Why? because that is the preferred way of writing radical fractions as far as i understand.

So we need to convert the sqrt%283%29%2Bsqrt%282%29 to a rational number. We do this by multiplying by 1 (so the thing we started with is unchanged but we write 1 in this example as +%28sqrt%283%29-sqrt%282%29%29%2F%28sqrt%283%29-sqrt%282%29%29+

So, we have

which is

Now what is the point of doing this? Well multiply out the terms on the denominator and they will simplify to 3-2 which in this case is a lovely 1.

+sqrt%283%29%28sqrt%283%29-sqrt%282%29%29%2F1+
+sqrt%283%29%28sqrt%283%29-sqrt%282%29%29+
+sqrt%283%29sqrt%283%29-sqrt%283%29sqrt%282%29+
+3-sqrt%283%29sqrt%282%29+
+3-sqrt%286%29+

As hopefully you can appreciate that this is somewhat simpler than what we started with.

Use your calculator to work out the value of both the original and my answer to show yourself that they are the same value.

Jon.