Mister Exam

x^2+4x+5 canonical form

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     2          
5 + x  + 4*x = 0
x2+4x+5=0x^{2} + 4 x + 5 = 0
x^2 + 4*x + 5 = 0
Detail solution
Given line equation of 2-order:
x2+4x+5=0x^{2} + 4 x + 5 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=1a_{11} = 1
a12=0a_{12} = 0
a13=2a_{13} = 2
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=5a_{33} = 5
To calculate the determinant
Δ=a11a12a12a22\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=1000\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|
Δ=0\Delta = 0
Because
Δ\Delta
is equal to 0, then
(x~+2)2=1\left(\tilde x + 2\right)^{2} = -1
x~2=1\tilde x'^{2} = -1
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
x~=x~+2\tilde x' = \tilde x + 2
y~=y~\tilde y' = \tilde y
The center of the canonical coordinate system in OXY
x0=x~cos(ϕ)y~sin(ϕ)x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}
y0=x~sin(ϕ)+y~cos(ϕ)y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}
x0=2+00x_{0} = -2 + 0 \cdot 0
y0=02y_{0} = - 0 \cdot 2
x0=2x_{0} = -2
y0=0y_{0} = 0
The center of canonical coordinate system at point O
(-2, 0)

Basis of the canonical coordinate system
e1=(1, 0)\vec e_1 = \left( 1, \ 0\right)
e2=(0, 1)\vec e_2 = \left( 0, \ 1\right)
Invariants method
Given line equation of 2-order:
x2+4x+5=0x^{2} + 4 x + 5 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=1a_{11} = 1
a12=0a_{12} = 0
a13=2a_{13} = 2
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=5a_{33} = 5
The invariants of the equation when converting coordinates are determinants:
I1=a11+a22I_{1} = a_{11} + a_{22}
     |a11  a12|
I2 = |        |
     |a12  a22|

I3=a11a12a13a12a22a23a13a23a33I_{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(λ)=a11λa12a12a22λI{\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
I1=1I_{1} = 1
     |1  0|
I2 = |    |
     |0  0|

I3=102000205I_{3} = \left|\begin{matrix}1 & 0 & 2\\0 & 0 & 0\\2 & 0 & 5\end{matrix}\right|
I(λ)=1λ00λI{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|
     |1  2|   |0  0|
K2 = |    | + |    |
     |2  5|   |0  5|

I1=1I_{1} = 1
I2=0I_{2} = 0
I3=0I_{3} = 0
I(λ)=λ2λI{\left(\lambda \right)} = \lambda^{2} - \lambda
K2=1K_{2} = 1
Because
I2=0I3=0K2>0I10I_{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 imaginary parallel lines
I1y~2+K2I1=0I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0
or
y~2+1=0\tilde y^{2} + 1 = 0
None

- reduced to canonical form