Olympiad math problem (An Irrational Equation )| Geendle

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In this video I solve an irrational equation by substitution.
#squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus, #matholympiad

[@echandler]

Your preview slide says 930 instead of 970. :).
a=5+sqrt(24)
b=5-sqrt(24)
a + b = 10
ab = 1
b=1/a
So a^x + 1/a^x = 970.
If x is a solution, so is -x.
We have
a + 1/a = 10
Squaring gives
a^2 + 1/a^2 = 100 - 2 = 98
Cubing gives
a^3 + 1/a^3 + 3(a+1/a) = 10^3
a^3 + 1/a^3 = 1000 - 30 = 970
Thus x = +3 or -3.
For positive values of x, a^x + b^x is increasing, thus there is only one solution there.
So this problem reduces to given a+b and ab, what is a^x + b^x for various values of x. Suppose x is an integer.

[@Geendle]

Hey friend. Thank you very much. I've already corrected the slide.
Your solution is very good. I will apply it in one of the next videos.