x³+64=0
If we find first zeroes of this cubic polynomial equation we have to replace X by negative 4
Since it is cubic equation so it has three roots
Imply -4 is the one root of this equation
Now X³= -64
Imply X = ± (-64)⅓
Imply X = ± i {(4)³}⅓
Imply X = 4i , – 4i
Thus the required roots of given equation are as :
4i , – 4i, -4
-4
-4
-4
-4
-4
-4
-4,-4w,-4w^2
-4,-4w,-4w^2
x=-4
X=-4
X1=-4, X2=2+(3.46410161514)i, X3=2-(3.46410161514)i
-4
-4
X=-4
-4
X=-4
-4
-4
X☞ -4
(x+4)(x^2-4x+16)=(x+4)(x-8)(x+4) sum between different cubes , third degree
-4×-4= 16
-4×-4×-4= -64
X³+64=0
X³=-64
X= -4
x³+64=0
If we find first zeroes of this cubic polynomial equation we have to replace X by negative 4
Since it is cubic equation so it has three roots
Imply -4 is the one root of this equation
Now X³= -64
Imply X = ± (-64)⅓
Imply X = ± i {(4)³}⅓
Imply X = 4i , – 4i
Thus the required roots of given equation are as :
4i , – 4i, -4
X³+64=0
X³= _64
X = _ 4
X= – 4 or 4i
-4:::::-4w:::::::-4w^2
-4
_4
4i
x=-4
X= -4
X=-4
3 soutions
X=-4
Undefined
x=-4
x=-4
x^3=(-4)^3
x=-4
x^3=-64
=-4^3
x=-4
X^3=-64
X=cube root of-64
X=-4
(x+4)(x^2-4x+16) = 0
Two complex roots
One real root: x = -4
Not funny!
Easy
X³+64=0
X³+64-64=0-64
X³=-64
X=-64³
X=-4
X³+64=0
X³=-64
X³=-(4)³
Cube cancel out the equation
X=-(4)
X=-4
✓ X=4i
+- 4i
There is no solution in Real Numbers!
x^3 = 164. The cube root of -64 is 4i
x^3+4^3=0.
(x+4)(x^2-4x+4^2)=0.
x=-4.
x^2-4x=-16.
(x-2)^2=-16+4=-12.
x-2=+/-(2i)(Sqrt 3).
x=2+/-(2i)(Sqrt 3).
X^3 + 4^3 = 0
Factor
(X + 4)(X^2 -4 X + 16) = 0
X = -4
Or
X^2 – 4 X +16 = 0
(X -2)^2 + 12 = 0
(X – 2)^2 = – 4(3)
= 4(3) i^2
X – 2 = +- 2 sqrt 3 i
X = 2 +- 2 sqrt 3 i
Where i = sqrt(-1)