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# help algebra

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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.

Feb 14, 2022

#1
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I dont understand your queston

Feb 14, 2022
#2
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$$(\sqrt{600} + \sqrt{150} + 4\sqrt{24})/(6\sqrt{32} - 3\sqrt{50} - \sqrt{2})*8 = a\sqrt{b}$$

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   Feb 14, 2022
edited by CPhill  Feb 14, 2022
#3
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Math is cool    Kakashi  Feb 14, 2022