Let x be a non-zero real number such that (x³+20x)^1/5=(x⁵-20x)^1/3. Find the product of all possible values of x.

         

LET  L=(x³+20x)^1/5  and
       R=(x⁵-20x)^1/3.     

Clearly L^5=(x³+20x)  and
           R^3=(x⁵-20x).  

So:  L^5+R^3=(x⁵-20x)+(x³+20x) 

  and   
      L^5+R^3=x⁵+x³

       R=L=x  is a solution (maybe not the only one)

 

SINCE   20*X=X^5-X^3  and x is not a zero 

  20=x^4-x^2=25-5

and x^2=5  is the answer.

BTW 

  X1= +sqrt(5);      X2=- sqrt(5);      real roots

    X3= sqrt(-4)=2i;     X4=- sqrt(-4)=-2i;  complex.

 Proving absence of other non-zero roots is  elementary (polynom of 5th degree!).

 

Nice puzzle    -  rated it 4.

Edited on February 24, 2014, 6:17 am

Posted by Ady TZIDON on 2014-02-24 06:07:03