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25x^3-10x^2+x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$25 x^{3} - 10 x^{2} + x = 0$$
transform
Take common factor $x$ from the equation
we get:
$$x \left(25 x^{2} - 10 x + 1\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$25 x^{2} - 10 x + 1 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 25$$
$$b = -10$$
$$c = 1$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 25 \cdot 4 \cdot 1 + \left(-10\right)^{2} = 0$$
Because D = 0, then the equation has one root.
$$x_{2} = \frac{1}{5}$$
The final answer for (25*x^3 - 10*x^2 + x) + 0 = 0:
$$x_{1} = 0$$
$$x_{2} = \frac{1}{5}$$
$$25 x^{3} - 10 x^{2} + x = 0$$
transform
Take common factor $x$ from the equation
we get:
$$x \left(25 x^{2} - 10 x + 1\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$25 x^{2} - 10 x + 1 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 25$$
$$b = -10$$
$$c = 1$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 25 \cdot 4 \cdot 1 + \left(-10\right)^{2} = 0$$
Because D = 0, then the equation has one root.
x = -b/2a = --10/2/(25)
$$x_{2} = \frac{1}{5}$$
The final answer for (25*x^3 - 10*x^2 + x) + 0 = 0:
$$x_{1} = 0$$
$$x_{2} = \frac{1}{5}$$
Vieta's Theorem
rewrite the equation
$$25 x^{3} - 10 x^{2} + x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - \frac{2 x^{2}}{5} + \frac{x}{25} = 0$$
$$p x^{2} + x^{3} + q x + v = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{2}{5}$$
$$q = \frac{c}{a}$$
$$q = \frac{1}{25}$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = \frac{2}{5}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = \frac{1}{25}$$
$$x_{1} x_{2} x_{3} = 0$$
$$25 x^{3} - 10 x^{2} + x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - \frac{2 x^{2}}{5} + \frac{x}{25} = 0$$
$$p x^{2} + x^{3} + q x + v = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{2}{5}$$
$$q = \frac{c}{a}$$
$$q = \frac{1}{25}$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = \frac{2}{5}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = \frac{1}{25}$$
$$x_{1} x_{2} x_{3} = 0$$
Sum and product of roots
[src]
sum
0 + 1/5
$$\left(0\right) + \left(\frac{1}{5}\right)$$
=
1/5
$$\frac{1}{5}$$
product
0 * 1/5
$$\left(0\right) * \left(\frac{1}{5}\right)$$
=
0
$$0$$
The graph