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 Penta-Quadratic (Posted on 2014-02-23)
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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 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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