Here's an approach without differentiating: By inspection, 1 is a root. Any other real root must be positive, negative, or zero. Apply Descartes' rule of signs to see that there can be no more than one positive real root, namely the 1. Substitute y=-x, rewrite the equation as y^2025 + 25 y + 26 = 0, and apply the rule of signs again to see there can be no positive real roots y of the rewritten equation, hence no negative real roots x of the original equation. 0 obviously does not solve the original equation, so the only real solution is x=1.
Modulars also we can use , basically 25x is congruent to 0 or 5 mod 10 , 26 congruent to 6 mod 10 , so 2 cases 1: 25x congruent to 0 mod 10 , which means x^2025 congruent to 6 mod 10 , which means x is congruent to 6 mod 10 which means x minimum has to be 6 which isnt possible It wont satisfy so this case eliminated 2 : 25x congruent to 5 mod 10 , which means x^2025 must be congruent to 1 mod 10 , which means x also must be congruent to 1 mod 10 , so x has to minimum be 1 , now this is clearly an increasing function therefore if x>1 the equation wont satisfy , so therefore x = 1 is the only solution (ok im assuming for integers solutions here , else by pure observation also only x = 1 works)
Here's an approach without differentiating: By inspection, 1 is a root. Any other real root must be positive, negative, or zero. Apply Descartes' rule of signs to see that there can be no more than one positive real root, namely the 1. Substitute y=-x, rewrite the equation as y^2025 + 25 y + 26 = 0, and apply the rule of signs again to see there can be no positive real roots y of the rewritten equation, hence no negative real roots x of the original equation. 0 obviously does not solve the original equation, so the only real solution is x=1.
Would there be 2024 non-real solutions?
Yes
Was that a “new-yearlial” equation?
My question too
Modulars also we can use , basically 25x is congruent to 0 or 5 mod 10 , 26 congruent to 6 mod 10 , so 2 cases
1: 25x congruent to 0 mod 10 , which means x^2025 congruent to 6 mod 10 , which means x is congruent to 6 mod 10 which means x minimum has to be 6 which isnt possible It wont satisfy so this case eliminated
2 : 25x congruent to 5 mod 10 , which means x^2025 must be congruent to 1 mod 10 , which means x also must be congruent to 1 mod 10 , so x has to minimum be 1 , now this is clearly an increasing function therefore if x>1 the equation wont satisfy , so therefore x = 1 is the only solution (ok im assuming for integers solutions here , else by pure observation also only x = 1 works)
There’s a guy (I think it is “super academy”) who does a maths channel. You need to train him how to make his material less sleepy (boring).
This applies to many of them. They don't realize that most of the time what we really need to know is the overall strategy that leads to the solution.
Sum of coefficients = 0 . . . x = 1
I got x=1 by inspection.
x = 1