If
(\sqrt{600} + \sqrt{150} + 4\sqrt{24})/(6\sqrt{32} - 3\sqrt{50} - \sqrt{2})*8 = a\sqrt{b},
where a and b are integers and b is as small as possible, find$a+b.
+128475 \((\sqrt{600} + \sqrt{150} + 4\sqrt{24})/(6\sqrt{32} - 3\sqrt{50} - \sqrt{2})*8 = a\sqrt{b}\)
Let's take this one term at a time
sqrt (600) = sqrt (100 * 6) = 10sqrt 6
sqrt (150 ) = sqrt ( 25 *6) = 5sqrt 6
4sqrt (24) = 4 sqrt ( 4 * 6) = 4*2 sqrt 6 = 8 sqrt 6
Adding these three terms we get 23 sqrt 6
Multiplying by 8 we get 184 sqrt 6
Now....look at the terms in the denominator
6sqrt ( 32) = 6sqrt ( 16 *2) = 6*4 sqrt ( 2) = 24sqrt 2
3sqrt (50) = 3 sqrt (25 * 2) = 3*5 sqrt (2) = 15sqrt 2
sqrt ( 2) = 1 * sqrt (2)
So the denominator = (24 - 15 - 1) sqrt 2 = 8sqrt 2
So we have 184 sqrt 6 / ( 8 sqrt 2) = (184/8) sqrt 6 /sqrt 2 = (23)sqrt 3
a + b = 23 + 3 = 26
