A Very Nice Math Olympiad Problem | Solve for real value of x for which x^30+x^20=80

[@GurnoorSingh-vx2nq]

this was an Olympiad prob???

[@michaeledwards2251]

Somebody loves 64 + 16 = 80. (x^10)^3 + (x^10)^2 = 80. Initially x^10 = 4, giving x = 2^(1/5) with another 9 roots to find.

[@RobertKvsv]

Explain?

[@michaeledwards2251]

@@RobertKvsv
Positive and Negative, and Complex, roots which raised to the 10th power evaluate to 4.

[@sy8146]

I would like to show the 2 points as follows:
1) There is another solution: x = - (4^(1/10)) [ There are 2 real solutions. ]
2) As for writing the final answer, 2^(1/5) is better than 4^(1/10) .
Therefore, my final answer is x = ± 2^(1/5) .
>

[@mikeeisler6463]

2 problems
1. 4^(1/10) isn’t simplest form
4^(1/10)
= (2^2)^(1/10)
= 2^(2/10)
= 2^(1/5)
2. There are 2 real solutions:
x^10 = 4
x^10 - 4 = 0
(x^5)^2 - 2^2 = 0
Difference of squares
(x^5 - 2)(x^5 + 2) = 0
x = +- 2^(1/5)

[@Berries1111_]

Or 10th root 4