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Solve (sqrt(x) - 1)*sqrt(x)*(sqrt(x) + 1) = 6.

 Dec 21, 2019
 #1
avatar+11725 
+2

(sqrt(x) - 1)*sqrt(x)*(sqrt(x) + 1)   = 6

                                               x   = 4

(sqrt(4) - 1)*sqrt(4)*(sqrt(4) + 1) = 6

                          (2 - 1)*2*(2 + 1) = 6

                                          1*2*3 = 6

laugh

 Dec 21, 2019
 #2
avatar+22096 
+1

Solve (sqrt(x) - 1)*sqrt(x)*(sqrt(x) + 1) = 6.

(sqrtx-1)*(x+sqrtx)

x sqrtx +x -x-sqrtx = 6

x sqrtx - sqrtx = 6

 

x^3/2 - x^1/2 = 6    let x = x^1/2  

x^3-x = 6

x(x^2-1) = 6

x(x-1)(x+1) = 6                               Maybe this: assuming x is positive,    factors of 6 are 1 2 3 6  

                                                           then you can deduce x = 2

                                                                 then since we substituted x = x^1/2   ,    x1/2 =2     or   x = 4

x1/2 (x1/2-1)(x1/2+1) = 6

    x = 4    (from wolframalpha....I did not know how to proceed)

 Dec 21, 2019
edited by ElectricPavlov  Dec 21, 2019
 #3
avatar+109563 
+1

( √x - 1) * √x * *( √x + 1)  = 6

 

Let  √x =  a  .....so we have

 

(a - 1)  * a  * ( a + 1)   =  6

 

(a - 1) (a + 1) a  =  6

 

(a^2 - 1) a  = 6

 

a^3  - a   = 6

 

a^3  - a  - 6   =  0         

 

By the rational roots theorem, the possible  values for a  = ± [ 1, 2, 3, 6 ]

 

By inspection     a =   2

 

Therefore   √x  =  2   ⇒    x  = 4

 

 

cool cool cool

 Dec 21, 2019

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