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