Mister Exam
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x³-2x²+x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x + \left(x^{3} - 2 x^{2}\right) = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(x^{2} - 2 x + 1\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$x^{2} - 2 x + 1 = 0$$
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
$$a = 1$$
$$b = -2$$
$$c = 1$$
, then
Because D = 0, then the equation has one root.
$$x_{2} = 1$$
The final answer for x^3 - 2*x^2 + x = 0:
$$x_{1} = 0$$
$$x_{2} = 1$$
$$x + \left(x^{3} - 2 x^{2}\right) = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(x^{2} - 2 x + 1\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$x^{2} - 2 x + 1 = 0$$
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
$$a = 1$$
$$b = -2$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (1) = 0
Because D = 0, then the equation has one root.
x = -b/2a = --2/2/(1)
$$x_{2} = 1$$
The final answer for x^3 - 2*x^2 + x = 0:
$$x_{1} = 0$$
$$x_{2} = 1$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = 1$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 2$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 1$$
$$x_{1} x_{2} x_{3} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = 1$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 2$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 1$$
$$x_{1} x_{2} x_{3} = 0$$