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- Equation:
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Equation 3y^2
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Equation (x+1)^2+(x-6)^2=2x^2
- Equation (cos2x+sinx)/sqrt(sin(x-pi/4))=0
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Equation (-x-5)(2x+4)=0
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- Integral of d{x}:
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3y^2
- Derivative of:
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3y^2
- The double integral of:
- 3y^2
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Identical expressions
- 3y^ two
- 3y squared
- 3y to the power of two
- 3y2
- 3y²
- 3y to the power of 2
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Similar expressions
- 2x^2+3y^2=3
3y^2 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
$$a = 3$$
$$b = 0$$
$$c = 0$$
, then
Because D = 0, then the equation has one root.
$$y_{1} = 0$$
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
$$a = 3$$
$$b = 0$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (3) * (0) = 0
Because D = 0, then the equation has one root.
y = -b/2a = -0/2/(3)
$$y_{1} = 0$$
Vieta's Theorem
rewrite the equation
$$3 y^{2} = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$y^{2} = 0$$
$$p y + y^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = 0$$
$$y_{1} y_{2} = 0$$
$$3 y^{2} = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$y^{2} = 0$$
$$p y + y^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = 0$$
$$y_{1} y_{2} = 0$$
The graph