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Similar expressions
- 4*x^3-19*x^2-19*x+6=0
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- 4*x^3-19*x^2+19*x-6=0
4*x^3-19*x^2+19*x+6=0 equation
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The solution
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3 2 4*x - 19*x + 19*x + 6 = 0
$$\left(19 x + \left(4 x^{3} - 19 x^{2}\right)\right) + 6 = 0$$
Detail solution
Given the equation:
$$\left(19 x + \left(4 x^{3} - 19 x^{2}\right)\right) + 6 = 0$$
transform
$$\left(19 x + \left(\left(- 19 x^{2} + \left(4 x^{3} - 32\right)\right) + 76\right)\right) - 38 = 0$$
or
$$\left(19 x + \left(\left(- 19 x^{2} + \left(4 x^{3} - 4 \cdot 2^{3}\right)\right) + 19 \cdot 2^{2}\right)\right) + \left(-19\right) 2 = 0$$
$$19 \left(x - 2\right) + \left(- 19 \left(x^{2} - 2^{2}\right) + 4 \left(x^{3} - 2^{3}\right)\right) = 0$$
$$19 \left(x - 2\right) + \left(- 19 \left(x - 2\right) \left(x + 2\right) + 4 \left(x - 2\right) \left(\left(x^{2} + 2 x\right) + 2^{2}\right)\right) = 0$$
Take common factor -2 + x from the equation
we get:
$$\left(x - 2\right) \left(\left(- 19 \left(x + 2\right) + 4 \left(\left(x^{2} + 2 x\right) + 2^{2}\right)\right) + 19\right) = 0$$
or
$$\left(x - 2\right) \left(4 x^{2} - 11 x - 3\right) = 0$$
then:
$$x_{1} = 2$$
and also
we get the equation
$$4 x^{2} - 11 x - 3 = 0$$
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
$$a = 4$$
$$b = -11$$
$$c = -3$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 3$$
$$x_{3} = - \frac{1}{4}$$
The final answer for 4*x^3 - 19*x^2 + 19*x + 6 = 0:
$$x_{1} = 2$$
$$x_{2} = 3$$
$$x_{3} = - \frac{1}{4}$$
$$\left(19 x + \left(4 x^{3} - 19 x^{2}\right)\right) + 6 = 0$$
transform
$$\left(19 x + \left(\left(- 19 x^{2} + \left(4 x^{3} - 32\right)\right) + 76\right)\right) - 38 = 0$$
or
$$\left(19 x + \left(\left(- 19 x^{2} + \left(4 x^{3} - 4 \cdot 2^{3}\right)\right) + 19 \cdot 2^{2}\right)\right) + \left(-19\right) 2 = 0$$
$$19 \left(x - 2\right) + \left(- 19 \left(x^{2} - 2^{2}\right) + 4 \left(x^{3} - 2^{3}\right)\right) = 0$$
$$19 \left(x - 2\right) + \left(- 19 \left(x - 2\right) \left(x + 2\right) + 4 \left(x - 2\right) \left(\left(x^{2} + 2 x\right) + 2^{2}\right)\right) = 0$$
Take common factor -2 + x from the equation
we get:
$$\left(x - 2\right) \left(\left(- 19 \left(x + 2\right) + 4 \left(\left(x^{2} + 2 x\right) + 2^{2}\right)\right) + 19\right) = 0$$
or
$$\left(x - 2\right) \left(4 x^{2} - 11 x - 3\right) = 0$$
then:
$$x_{1} = 2$$
and also
we get the equation
$$4 x^{2} - 11 x - 3 = 0$$
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
$$a = 4$$
$$b = -11$$
$$c = -3$$
, then
D = b^2 - 4 * a * c =
(-11)^2 - 4 * (4) * (-3) = 169
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 3$$
$$x_{3} = - \frac{1}{4}$$
The final answer for 4*x^3 - 19*x^2 + 19*x + 6 = 0:
$$x_{1} = 2$$
$$x_{2} = 3$$
$$x_{3} = - \frac{1}{4}$$
Vieta's Theorem
rewrite the equation
$$\left(19 x + \left(4 x^{3} - 19 x^{2}\right)\right) + 6 = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - \frac{19 x^{2}}{4} + \frac{19 x}{4} + \frac{3}{2} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{19}{4}$$
$$q = \frac{c}{a}$$
$$q = \frac{19}{4}$$
$$v = \frac{d}{a}$$
$$v = \frac{3}{2}$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = \frac{19}{4}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = \frac{19}{4}$$
$$x_{1} x_{2} x_{3} = \frac{3}{2}$$
$$\left(19 x + \left(4 x^{3} - 19 x^{2}\right)\right) + 6 = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - \frac{19 x^{2}}{4} + \frac{19 x}{4} + \frac{3}{2} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{19}{4}$$
$$q = \frac{c}{a}$$
$$q = \frac{19}{4}$$
$$v = \frac{d}{a}$$
$$v = \frac{3}{2}$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = \frac{19}{4}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = \frac{19}{4}$$
$$x_{1} x_{2} x_{3} = \frac{3}{2}$$
Rapid solution
[src]
x1 = -1/4
$$x_{1} = - \frac{1}{4}$$
x2 = 2
$$x_{2} = 2$$
x3 = 3
$$x_{3} = 3$$
x3 = 3
Sum and product of roots
[src]
sum
2 - 1/4 + 3
$$\left(- \frac{1}{4} + 2\right) + 3$$
=
19/4
$$\frac{19}{4}$$
product
2*(-1) ------*3 4
$$3 \frac{\left(-1\right) 2}{4}$$
=
-3/2
$$- \frac{3}{2}$$
-3/2