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x3+640=0?
If that's the case, then you first want to move the 640 to the other side.
x3=-640
Then you take the cube root of -640 to find what x is. Or you can also write it as x=(-640) to the 1/3 power.
x=(-640)1/3
Which is equal to 4*101/3 or ≈(-8.618).
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∙ 12y ago
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How do you factorise x3 plus 2x2-8x plus 5 equals 0?
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
What is the solution to x3 plus 125 equals 0?
It is x = -5
How do you solve x3 plus 125 is equals to 0?
Realising that 125 is a perfect cube, this is actually pretty easy: x3 + 125 = 0 x3 = -125 x = -5
X3-11x2 plus 18x equals 0?
x3 - 11x2 + 18x = 0x (x2 - 11x + 18) = 0 . . . . x = 0 is a root.x (x - 9) ( x - 2) = 0 . . . . x = 9 and x = 2 are roots.X = 0, 2, 9
What are the points of intersection of y equals x squared plus 2x plus 2 with y equals x cubed plus 2?
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
How do you factorise x3 plus 2x2-8x plus 5 equals 0?
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
What is the solution to x3 plus 125 equals 0?
It is x = -5
How do you solve x3 plus 125 is equals to 0?
Realising that 125 is a perfect cube, this is actually pretty easy: x3 + 125 = 0 x3 = -125 x = -5
What is 0 plus x cubed?
Anything plus 0 is itself, so x3 + 0 is x3.
How many complex roots does 6x4 plus x3 - 5 equals 0 have?
It has two complex roots.
X3-11x2 plus 18x equals 0?
x3 - 11x2 + 18x = 0x (x2 - 11x + 18) = 0 . . . . x = 0 is a root.x (x - 9) ( x - 2) = 0 . . . . x = 9 and x = 2 are roots.X = 0, 2, 9
What is the trinomial of X3 plus 2x?
It could be x3 + 2x + 0It could be x3 + 2x + 0It could be x3 + 2x + 0It could be x3 + 2x + 0
What is x if x cubed plus eight equals zero?
x3 + 8 = 0, then x3 = -8, therefore x = -2 -2 x -2 = 4 4 x -2 = -8
What are the points of intersection of y equals x squared plus 2x plus 2 with y equals x cubed plus 2?
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
How do you solve x3 plus x2-x equals 0?
x3 + x2 - x = 0 ∴ x2 + x - 1 = 0 ∴ x2 + x + 0.25 = 1.25 ∴ (x + 0.5)2 = 1.25 ∴ x + 0.5 = ± 1.251/2 ∴ x = -0.5 ± 1.251/2
What are the real roots of x5 - 2x4 - 24x3 equals 0?
x5 - 2x4 - 24x3 = 0 x3(x2 - 2x - 24) = 0 x3(x2 + 4x - 6x - 24) = 0 x3(x - 6)(x + 4) = 0 x = {-4, 0, 6}
How do you solve x³-4x equals 0 algebraically?
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
