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Similar expressions
- x^3+x^2-4*x+4=0
- x^3-x^2-4*x-4=0
- x^3-x^2+4*x+4=0
x^3-x^2-4*x+4=0 equation
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3 2 x - x - 4*x + 4 = 0
$$\left(- 4 x + \left(x^{3} - x^{2}\right)\right) + 4 = 0$$
Detail solution
Given the equation:
$$\left(- 4 x + \left(x^{3} - x^{2}\right)\right) + 4 = 0$$
transform
$$\left(- 4 x + \left(\left(- x^{2} + \left(x^{3} - 1\right)\right) + 1\right)\right) + 4 = 0$$
or
$$\left(- 4 x + \left(\left(- x^{2} + \left(x^{3} - 1^{3}\right)\right) + 1^{2}\right)\right) + 4 = 0$$
$$- 4 \left(x - 1\right) + \left(- (x^{2} - 1^{2}) + \left(x^{3} - 1^{3}\right)\right) = 0$$
$$- 4 \left(x - 1\right) + \left(- (x - 1) \left(x + 1\right) + \left(x - 1\right) \left(\left(x^{2} + x\right) + 1^{2}\right)\right) = 0$$
Take common factor -1 + x from the equation
we get:
$$\left(x - 1\right) \left(\left(- (x + 1) + \left(\left(x^{2} + x\right) + 1^{2}\right)\right) - 4\right) = 0$$
or
$$\left(x - 1\right) \left(x^{2} - 4\right) = 0$$
then:
$$x_{1} = 1$$
and also
we get the equation
$$x^{2} - 4 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = -4$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 2$$
$$x_{3} = -2$$
The final answer for x^3 - x^2 - 4*x + 4 = 0:
$$x_{1} = 1$$
$$x_{2} = 2$$
$$x_{3} = -2$$
$$\left(- 4 x + \left(x^{3} - x^{2}\right)\right) + 4 = 0$$
transform
$$\left(- 4 x + \left(\left(- x^{2} + \left(x^{3} - 1\right)\right) + 1\right)\right) + 4 = 0$$
or
$$\left(- 4 x + \left(\left(- x^{2} + \left(x^{3} - 1^{3}\right)\right) + 1^{2}\right)\right) + 4 = 0$$
$$- 4 \left(x - 1\right) + \left(- (x^{2} - 1^{2}) + \left(x^{3} - 1^{3}\right)\right) = 0$$
$$- 4 \left(x - 1\right) + \left(- (x - 1) \left(x + 1\right) + \left(x - 1\right) \left(\left(x^{2} + x\right) + 1^{2}\right)\right) = 0$$
Take common factor -1 + x from the equation
we get:
$$\left(x - 1\right) \left(\left(- (x + 1) + \left(\left(x^{2} + x\right) + 1^{2}\right)\right) - 4\right) = 0$$
or
$$\left(x - 1\right) \left(x^{2} - 4\right) = 0$$
then:
$$x_{1} = 1$$
and also
we get the equation
$$x^{2} - 4 = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = -4$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (-4) = 16
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 2$$
$$x_{3} = -2$$
The final answer for x^3 - x^2 - 4*x + 4 = 0:
$$x_{1} = 1$$
$$x_{2} = 2$$
$$x_{3} = -2$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -1$$
$$q = \frac{c}{a}$$
$$q = -4$$
$$v = \frac{d}{a}$$
$$v = 4$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 1$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -4$$
$$x_{1} x_{2} x_{3} = 4$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -1$$
$$q = \frac{c}{a}$$
$$q = -4$$
$$v = \frac{d}{a}$$
$$v = 4$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 1$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -4$$
$$x_{1} x_{2} x_{3} = 4$$
Sum and product of roots
[src]
sum
-2 + 1 + 2
$$\left(-2 + 1\right) + 2$$
=
1
$$1$$
product
-2*2
$$- 4$$
=
-4
$$-4$$
-4