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- 2y-0.5y^2=0
2y+0.5y^2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = \frac{1}{2}$$
$$b = 2$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$y_{1} = 0$$
$$y_{2} = -4$$
a*y^2 + b*y + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = \frac{1}{2}$$
$$b = 2$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(2)^2 - 4 * (1/2) * (0) = 4
Because D > 0, then the equation has two roots.
y1 = (-b + sqrt(D)) / (2*a)
y2 = (-b - sqrt(D)) / (2*a)
or
$$y_{1} = 0$$
$$y_{2} = -4$$
Vieta's Theorem
rewrite the equation
$$\frac{y^{2}}{2} + 2 y = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$y^{2} + 4 y = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 4$$
$$q = \frac{c}{a}$$
$$q = 0$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = -4$$
$$y_{1} y_{2} = 0$$
$$\frac{y^{2}}{2} + 2 y = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$y^{2} + 4 y = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 4$$
$$q = \frac{c}{a}$$
$$q = 0$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = -4$$
$$y_{1} y_{2} = 0$$