Mister Exam
9-y² 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
$$a = -1$$
$$b = 0$$
$$c = 9$$
, then
Because D > 0, then the equation has two roots.
or
$$y_{1} = -3$$
$$y_{2} = 3$$
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
$$a = -1$$
$$b = 0$$
$$c = 9$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (-1) * (9) = 36
Because D > 0, then the equation has two roots.
y1 = (-b + sqrt(D)) / (2*a)
y2 = (-b - sqrt(D)) / (2*a)
or
$$y_{1} = -3$$
$$y_{2} = 3$$
Vieta's Theorem
rewrite the equation
$$9 - y^{2} = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$y^{2} - 9 = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -9$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = 0$$
$$y_{1} y_{2} = -9$$
$$9 - y^{2} = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$y^{2} - 9 = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -9$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = 0$$
$$y_{1} y_{2} = -9$$