Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation (-5*x-3)*(2*x-1)=0
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Equation 9*(x-5)=-x
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Equation 10*(x-9)=7
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Equation -x=8
- Express {x} in terms of y where:
- -10*x+16*y=-6
- -11*x-14*y=-3
- 4*x-1*y=-3
- -10*x-18*y=15
- Derivative of:
- 4*x^3-16*x
-
Identical expressions
- four *x^ three - sixteen *x= zero
- 4 multiply by x cubed minus 16 multiply by x equally 0
- four multiply by x to the power of three minus sixteen multiply by x equally zero
- 4*x3-16*x=0
- 4*x³-16*x=0
- 4*x to the power of 3-16*x=0
- 4x^3-16x=0
- 4x3-16x=0
- 4*x^3-16*x=O
-
Similar expressions
- 4*x^3+16*x=0
4*x^3-16*x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$4 x^{3} - 16 x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(4 x^{2} - 16\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$4 x^{2} - 16 = 0$$
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
$$a = 4$$
$$b = 0$$
$$c = -16$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 2$$
$$x_{3} = -2$$
The final answer for 4*x^3 - 16*x = 0:
$$x_{1} = 0$$
$$x_{2} = 2$$
$$x_{3} = -2$$
$$4 x^{3} - 16 x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(4 x^{2} - 16\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$4 x^{2} - 16 = 0$$
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
$$a = 4$$
$$b = 0$$
$$c = -16$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (4) * (-16) = 256
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 4*x^3 - 16*x = 0:
$$x_{1} = 0$$
$$x_{2} = 2$$
$$x_{3} = -2$$
Vieta's Theorem
rewrite the equation
$$4 x^{3} - 16 x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - 4 x = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -4$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -4$$
$$x_{1} x_{2} x_{3} = 0$$
$$4 x^{3} - 16 x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - 4 x = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -4$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -4$$
$$x_{1} x_{2} x_{3} = 0$$
Sum and product of roots
[src]
sum
-2 + 2
$$-2 + 2$$
=
0
$$0$$
product
-2*0*2
$$2 \left(- 0\right)$$
=
0
$$0$$
0
The graph