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Penta-Quadratic (Posted on 2014-02-23) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (1 votes)

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Solution my way | Comment 2 of 5 |

         

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

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