Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 9^x-8*3^x-9=0
- Equation x+y=4
-
Equation 5x+8=-12
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Equation x/9+x=-3
- Express {x} in terms of y where:
- 18*x+17*y=-14
- 16*x-17*y=-15
- -4*x-4*y=19
- 15*x+1*y=19
- Graphing y =:
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x^3-3x+2
- Derivative of:
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x^3-3x+2
- Integral of d{x}:
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x^3-3x+2
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Identical expressions
- x^ three -3x+ two = zero
- x cubed minus 3x plus 2 equally 0
- x to the power of three minus 3x plus two equally zero
- x3-3x+2=0
- x³-3x+2=0
- x to the power of 3-3x+2=0
- x^3-3x+2=O
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Similar expressions
- x^3+3x+2=0
- x^3-3x-2=0
x^3-3x+2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(x^{3} - 3 x\right) + 2 = 0$$
transform
$$\left(- 3 x + \left(x^{3} - 1\right)\right) + 3 = 0$$
or
$$\left(- 3 x + \left(x^{3} - 1^{3}\right)\right) + 3 = 0$$
$$- 3 \left(x - 1\right) + \left(x^{3} - 1^{3}\right) = 0$$
$$\left(x - 1\right) \left(\left(x^{2} + x\right) + 1^{2}\right) - 3 \left(x - 1\right) = 0$$
Take common factor -1 + x from the equation
we get:
$$\left(x - 1\right) \left(\left(\left(x^{2} + x\right) + 1^{2}\right) - 3\right) = 0$$
or
$$\left(x - 1\right) \left(x^{2} + x - 2\right) = 0$$
then:
$$x_{1} = 1$$
and also
we get the equation
$$x^{2} + x - 2 = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = -2$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 1$$
$$x_{3} = -2$$
The final answer for x^3 - 3*x + 2 = 0:
$$x_{1} = 1$$
$$x_{2} = 1$$
$$x_{3} = -2$$
$$\left(x^{3} - 3 x\right) + 2 = 0$$
transform
$$\left(- 3 x + \left(x^{3} - 1\right)\right) + 3 = 0$$
or
$$\left(- 3 x + \left(x^{3} - 1^{3}\right)\right) + 3 = 0$$
$$- 3 \left(x - 1\right) + \left(x^{3} - 1^{3}\right) = 0$$
$$\left(x - 1\right) \left(\left(x^{2} + x\right) + 1^{2}\right) - 3 \left(x - 1\right) = 0$$
Take common factor -1 + x from the equation
we get:
$$\left(x - 1\right) \left(\left(\left(x^{2} + x\right) + 1^{2}\right) - 3\right) = 0$$
or
$$\left(x - 1\right) \left(x^{2} + x - 2\right) = 0$$
then:
$$x_{1} = 1$$
and also
we get the equation
$$x^{2} + x - 2 = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = -2$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (1) * (-2) = 9
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 1$$
$$x_{3} = -2$$
The final answer for x^3 - 3*x + 2 = 0:
$$x_{1} = 1$$
$$x_{2} = 1$$
$$x_{3} = -2$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -3$$
$$v = \frac{d}{a}$$
$$v = 2$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -3$$
$$x_{1} x_{2} x_{3} = 2$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -3$$
$$v = \frac{d}{a}$$
$$v = 2$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -3$$
$$x_{1} x_{2} x_{3} = 2$$
The graph