A Special Cubic Equation
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- Опубликовано: 7 фев 2025
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x^2-x^3=80
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whoops you put (y+½)² at 1:59
About the 1st path: you cannot substitute x as y+1/3 and y+1/2 at the same time in the same equation.
0= x^2 - x^3 - 80 = x^2 - 16 - (x^3 + 64) = (x+4)(x-4) - (x+4)(x^2 -4x + 16) -->
(x+4)(x^2 - 5x + 20) = 0 ...
x^3-x^2+/-x+80=0 , (x+4)(x^2-5x+20)=0 , x= -4 , x=(5+/-V(25-80))/2 , x=(5+/-i*V(55))/2 ,
1 4
-5 -20
20 80
Cubic with trivial root, can be solved in memory
Χ=-4,χ=(5+-55^(1/2)*i)/2.
Other Method. x^3-x^2+80=0. Obvius root x=(-4) since (-4)^3-(-4)^2+80=-64-16+80=0. We apply Horner's Short Division x=-4
1 -1 0 80 || -4
-4 +20 -80
1 -5 20 0
So (x-(-4))•(x^2-5x+20)=0 so x+4=0 or x=(+5+-√-55)/2 so x1=-4 or x2,3=(5+-i•√55)/2.
x = -4
x^3-x^2=-80
(x+4)(x^2-5x+20)=0
x = {5+-(25-80)^(1/2)}/2 = [5+-{5^(1/2)*11^(1/2)}i]*2^(-1)
Got em all using RRT and long division.
x=-4,x=(5+i√55)/2,x=(5-i√55)/2
By a special Syberguy.
x^3-x^2+80=0 so x^3+64-x^2+16=0 so (x^3+4^3)-(x^2-4^2)=0 so (x+4)(x^2-4x+16)-(x+4)(x-4)=0 so (x+4)(x^2-4x+16-x+4)=0 so (x+4)(x^2-5x+20)=0 so x+4=0 or x^2-5x+4=0 so x1=-4 or x2,3=(+5+-√25-80)/2=(5+-i•√55)/2.
The third summand is 20(x+4) not 4(x+4), in your second method.
Exactly
-4 ?
Yes.
La met.1 x^2=(y+1/3) nu (y+ 1/2) in rest totul ok
you are mistaken