Japanese | Can you solve this ? | Nice Math Olympiad Problem
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- เผยแพร่เมื่อ 5 เม.ย. 2024
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X^3 = (x-4)^3. Or. - x^3 =(x-4,)^3
Now quadratic equation and second we get X=2 solution and another ( x-2)* ( quadratic equation ) to solve
It would be very good if you show all the operations in a square after making the trades.
All operations are given in the last 10 seconds in a photo.
In case you didn't see my last comment, you shouldn't use the x symbol for multiplication, because then it will be confused with the variable x
Learn these tricks and quickly solve for all five roots. Substitute x=y+2 into the given equation and rearrange to (y+2)^6-(y-2)^6=0
Pascal's Triangle 1 6 15 20 15 6 1 Even powers of y cancel out and odd powers double. Thus, 2[6*2^1*y^5+20*2^3*y^3+6*2^5y]=0
8y[3y^4+40y^2+48]=0; y=0 or y^2=(-40±32)/6=-4/3 or -12. Thus, y=0, ±i2√3/3, or ±i2√3 and x=y+2=2, 2±i2√3/3, or 2±i2√3
Check: 2^6-(-2)^6=0; arg(±2±i2√3/3)=±30° and arg(±2±i2√3)=±60°. 6(±30°)=±180° and 6(±60°)=±360°, so x^6-(x-4)^6=0
Only possible when x = -(x-4)
Solve: x= 2
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Given equation is of 6th degree, hence it should have six roots. You have shown only 5 roots.
Estimado Ecotú. Si no te complicás, no sos vos.
Решил в уме за 30 сек. х=2
Я за 2 😂
Here
x^6 = ( x -4 )^6
It means
x - 4 = - x
2x -4 = 0
2x = 4
x = 2
Why waste 16 minutes .
That's right.
To be formal, you take the 6th rooth of both sides, which yields 3 scenarios.
1. X=x-4
2. -x=x-4
3. X=-(x-4)
Only solution for these is 2.
I hate people like this. He is finding the OTHER solutions as well??? Do you actually not understand that?