+590 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.
+129899 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
