+0  
 
+1
107
4
avatar

If\[\displaystyle\frac{\sqrt{600} + \sqrt{150} + 4\sqrt{54}}{6\sqrt{32} - 3\sqrt{50} - \sqrt{72}} = a\sqrt{b},\]where $a$ and $b$ are integers and $b$ is as small as possible, find $a+b.$

 Jul 13, 2021
 #1
avatar
+1

Let's simplify: (10√6 + 5√6 + 12√6)     /       (24√2 - 15√2 - 6√2)      

Can you take it from here? 

 Jul 13, 2021
 #2
avatar+121065 
+2

sqrt 600  =   10sqrt 6

sqrt 150  =    5sqrt 6

4 sqrt 54   =   4*3 * sqrt (6)  = 12sqrt 6

So.....the  numerator =  27sqrt 6

 

6sqrt 32  =    6 * 4 sqrt 2  =      24 sqrt 2

-3sqrt 50  =   - 5*3 sqrt 2 =       -  15 sqrt 2

-sqrt 72    =    - 6sqrt (2)

So.....the  denominator =  3sqrt 2

 

So

 

27sqrt 6

_______  =       9  sqrt  ( 6/2)    =      3  sqrt 3

3 sqrt 2

 

cool cool cool

 Jul 13, 2021
 #4
avatar+8 
0

Check - It's incorrect.

fuvtest1  Jul 18, 2021
 #3
avatar+8 
+1

$\sqrt{600}=\sqrt{6}\cdot\sqrt{100}=10\sqrt{6}$

$\sqrt{150}=\sqrt{25}\cdot\sqrt{6}=5\sqrt{6}$

$4\sqrt{54}=4{\sqrt{9}\cdot\sqrt{6}}=(4\cdot3)\sqrt{6}=12\sqrt{6}$

 

$6\sqrt{32}=6(\sqrt{16}\cdot\sqrt{2})=(6\cdot4)\sqrt{2}=24\sqrt{2}$

$3\sqrt{50}=3(\sqrt{25}\cdot\sqrt{2})=(3*5)\sqrt{2}=15\sqrt{2}$

$\sqrt{72}=\sqrt{36}\cdot\sqrt{2}=6\sqrt{2}$

 

$\frac{10\sqrt{6}+5\sqrt{6}+12\sqrt{6}}{24\sqrt{2}-15\sqrt{2}-6\sqrt{2}}=\frac{27\sqrt{6}}{3\sqrt{2}}=\frac{27}{3}\cdot\sqrt{\frac{6}{2}}=9\sqrt{3}$

 

$9+3=12$

12

 Jul 16, 2021

38 Online Users

avatar