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Hey guys, please help me answer a question.

 

Solve the equation

*sqrt3 is cube root

 

sqrt3( 60 - x) + sqrt3 ( x- 11) = sqrt3 (4)

 

Plus, can you guys give some hints on simplifying cube root radicals?

Thanks a lot guys, kind of struggling here. 

 Aug 15, 2019
 #1
avatar+103122 
+1

This one is a little difficult....but....a substitution later on will help us to turn it into a quadratic

 

∛(60-x)  + ∛(x - 11)  = ∛4       cube both sides and we get that

 

49  + 3∛(x-11)∛[60-x)^2  + 3∛(60-x)∛(x-11)^2  =  4

 

3∛(x-11)∛(60-x)^2  + 3∛(60-x)∛(x-11)^2  =  -45

 

3 [ ∛(x-11)∛(60-x)^2  + ∛(60-x)∛(x-11)^2]  = -45      divide both sides  by  3  and factor the left side

 

[ ∛(x-11)∛(60-x) ] [ ∛(60-x)  + ∛(x - 11)  ]  =  -15

 

∛[(60-x) (x - 11) ] [ ∛(60-x)  + ∛(x - 11)  ]  =  -15     factor out -1

 

∛[ (-1)(x - 60)(x - 11) ]  [ ∛(60-x)  + ∛(x - 11)  ]  =  -15

 

(-1) ∛[ (x - 60)(x - 11) ]  [ ∛(60-x)  + ∛(x - 11)  ]  =  -15      divide both sides  by -1

 

∛[ (x - 60)(x - 11) ]  [ ∛(60-x)  + ∛(x - 11)  ]  =  15     

 

 Since   ∛(60-x)  + ∛(x - 11)  = ∛4     we have that

 

∛[ (x - 60)(x - 11) ] [  ∛4 ]  =  15      cube both sides again

 

(x - 60)(x - 11)  (4)  = 15^3

 

4 ( x^2  - 71x  + 660) = 3375

 

4x^2  - 284x  + 2640 = 3375    subtract 3375 from  both sides

 

4x^2  - 284x - 735   = 0        factor ( believe it or not)  as

 

(2x - 147) ( 2x + 5)   = 0      set both factors to  0   and solve for  x

 

2x - 147  = 0             2x  + 5  =0

2x  = 147                  2x  = -5

x =147/2                     x  = -5/2

 

 

 

cool cool cool

 Aug 15, 2019
edited by CPhill  Aug 15, 2019

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