Mister Exam
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Identical expressions
- 16x^ three -x= zero
- 16x cubed minus x equally 0
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Similar expressions
- 16x^3+x=0
16x^3-x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$16 x^{3} - x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(16 x^{2} - 1\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$16 x^{2} - 1 = 0$$
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
$$a = 16$$
$$b = 0$$
$$c = -1$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = \frac{1}{4}$$
Simplify
$$x_{3} = - \frac{1}{4}$$
Simplify
The final answer for (16*x^3 - x) + 0 = 0:
$$x_{1} = 0$$
$$x_{2} = \frac{1}{4}$$
$$x_{3} = - \frac{1}{4}$$
$$16 x^{3} - x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(16 x^{2} - 1\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$16 x^{2} - 1 = 0$$
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
$$a = 16$$
$$b = 0$$
$$c = -1$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (16) * (-1) = 64
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = \frac{1}{4}$$
Simplify
$$x_{3} = - \frac{1}{4}$$
Simplify
The final answer for (16*x^3 - x) + 0 = 0:
$$x_{1} = 0$$
$$x_{2} = \frac{1}{4}$$
$$x_{3} = - \frac{1}{4}$$
Vieta's Theorem
rewrite the equation
$$16 x^{3} - x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - \frac{x}{16} = 0$$
$$p x^{2} + x^{3} + q x + v = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = - \frac{1}{16}$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = - \frac{1}{16}$$
$$x_{1} x_{2} x_{3} = 0$$
$$16 x^{3} - x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - \frac{x}{16} = 0$$
$$p x^{2} + x^{3} + q x + v = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = - \frac{1}{16}$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = - \frac{1}{16}$$
$$x_{1} x_{2} x_{3} = 0$$
Sum and product of roots
[src]
sum
0 - 1/4 + 0 + 1/4
$$\left(\left(- \frac{1}{4} + 0\right) + 0\right) + \frac{1}{4}$$
=
0
$$0$$
product
1*-1/4*0*1/4
$$1 \left(- \frac{1}{4}\right) 0 \cdot \frac{1}{4}$$
=
0
$$0$$
0
Rapid solution
[src]
x1 = -1/4
$$x_{1} = - \frac{1}{4}$$
x2 = 0
$$x_{2} = 0$$
x3 = 1/4
$$x_{3} = \frac{1}{4}$$
The graph