+0

0
207
1
+581

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
+109560
+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

Aug 15, 2019
edited by CPhill  Aug 15, 2019