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Canonical form/ x^2+5x

x^2+5x canonical form

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The solution

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x  + 5*x = 0
x2+5x=0x^{2} + 5 x = 0 
x^2 + 5*x = 0
Detail solution
Given line equation of 2-order:
x2+5x=0x^{2} + 5 x = 0
This equation looks like:
a11x2+2a12xy+a22y2+2a13x+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0
where
a11=1a_{11} = 1
a12=0a_{12} = 0
a13=52a_{13} = \frac{5}{2}
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=0a_{33} = 0
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~+52)2=254\left(\tilde x + \frac{5}{2}\right)^{2} = \frac{25}{4}
x~′2=254\tilde x'^{2} = \frac{25}{4}
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
x~′=x~+52\tilde x' = \tilde x + \frac{5}{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=(−52)1+0⋅0x_{0} = \left(- \frac{5}{2}\right) 1 + 0 \cdot 0
y0=(−52)0+0⋅1y_{0} = \left(- \frac{5}{2}\right) 0 + 0 \cdot 1
x0=−52x_{0} = - \frac{5}{2}
y0=0y_{0} = 0
The center of canonical coordinate system at point O
(-5/2, 0)

Basis of the canonical coordinate system
e⃗1=(1, 0)\vec e_{1} = \left( 1, \ 0\right)
e⃗2=(0, 1)\vec e_{2} = \left( 0, \ 1\right)
Invariants method
Given line equation of 2-order:
x2+5x=0x^{2} + 5 x = 0
This equation looks like:
a11x2+2a12xy+a22y2+2a13x+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0
where
a11=1a_{11} = 1
a12=0a_{12} = 0
a13=52a_{13} = \frac{5}{2}
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=0a_{33} = 0
The invariants of the equation when converting coordinates are determinants:
I1=a11+a22I_{1} = a_{11} + a_{22}
     |a11  a12|
I2 = |        |
     |a12  a22|

I3=∣a11a12a13a12a22a23a13a23a33∣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(λ)=∣a11−λa12a12a22−λ∣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
I1=1I_{1} = 1
     |1  0|
I2 = |    |
     |0  0|

I3=∣10520005200∣I_{3} = \left|\begin{matrix}1 & 0 & \frac{5}{2}\\0 & 0 & 0\\\frac{5}{2} & 0 & 0\end{matrix}\right|
I(λ)=∣−λ+100−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 1 & 0\\0 & - \lambda\end{matrix}\right|
     | 1   5/2|   |0  0|
K2 = |        | + |    |
     |5/2   0 |   |0  0|

I1=1I_{1} = 1
I2=0I_{2} = 0
I3=0I_{3} = 0
I(λ)=λ2−λI{\left(\lambda \right)} = \lambda^{2} - \lambda
K2=−254K_{2} = - \frac{25}{4}
Because
I2=0∧I3=0∧K2<0∧I1≠0I_{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
I1y~2+K2I1=0I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0
or
y~2−254=0\tilde y^{2} - \frac{25}{4} = 0
None

- reduced to canonical form
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