Best Answer

x^(3) +x^(2) - x = 0

Factor

x(x^(2) + x - 1) = 0

First part of the answerr is x = 0

Second part of the answer

x^(2) + x - 1 = 0

Use Quadratic Eq'n.

x = { -1 +/- sqrt[1^(2) - 4(1)(-1_]} / 2(1)

x = { -1 +/- sqrt[1 + 4]} / 2

x = ( -1 +/- sqrt)5)} / 2

x = =1 +/- 2.236...} / 2

x = -1/2 +/- 2.236.../2

X = -0.5 +/- 1.118...

x = - 1.618.... & x = 0.618....

So the three values of x that satisfy the equ'n are

x = 0 , -1.618.... & 0.618....

Q: How do you solve x3 plus x2-x equals 0?

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x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)

IMPOSSIBLE

x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.

No.

It is x = -5

Related questions

x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)

IMPOSSIBLE

cube root of 216 = 6 > X= 6

x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.

No.

y2=x3+3x2

No, it is not.

It is x = -5

(xn+2-1)/(x2-1)

x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4

x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0

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.