Mister Exam

-2x^2+3x+5 canonical form

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       2          
5 - 2*x  + 3*x = 0
$$- 2 x^{2} + 3 x + 5 = 0$$
-2*x^2 + 3*x + 5 = 0
Detail solution
Given line equation of 2-order:
$$- 2 x^{2} + 3 x + 5 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-2 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$- 2 x^{2} + 3 x + 5 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 5$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
     |a11  a12|
I2 = |        |
     |a12  a22|

$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
     |a11  a13|   |a22  a23|
K2 = |        | + |        |
     |a13  a33|   |a23  a33|

substitute coefficients
$$I_{1} = -2$$
     |-2  0|
I2 = |     |
     |0   0|

$$I_{3} = \left|\begin{matrix}-2 & 0 & \frac{3}{2}\\0 & 0 & 0\\\frac{3}{2} & 0 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 2 & 0\\0 & - \lambda\end{matrix}\right|$$
     |-2   3/2|   |0  0|
K2 = |        | + |    |
     |3/2   5 |   |0  5|

$$I_{1} = -2$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 2 \lambda$$
$$K_{2} = - \frac{49}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$\frac{49}{8} - 2 \tilde y^{2} = 0$$
None

- reduced to canonical form