Mister Exam
Other calculators
- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
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- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
- 1/2x^2+2x^2+3x^2-x1x2+x1x3-x2x3
- 3*(x-2)*(x+2)
- 16x^2+0xy-25y^2-160x-100y-100=0
- 4*x^2+4*x*y+8*x*z-3*y^2+4*z^2
- Plot:
- x^2+(y-sqrt((|x|)))^2=1
- z=3*x^2+2*y^2+x*y-5*x+y-1
- z=-3*x^2+6*x-3*y^2-6*y
-
tg(x^100+y^100)=0
- Integral of d{x}:
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(x+2)(x-3)
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Identical expressions
- (x+ two)(x- three)
- (x plus 2)(x minus 3)
- (x plus two)(x minus three)
- x+2x-3
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Similar expressions
- (3x+2)*(x-3)
- (x-2)(x-3)
- 2ln(x)+2*(x)-3
- f=5x+2x-3x
- cos(x)+2*x-3=0
- (x+2)(x+3)
(x+2)(x-3) canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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(-3 + x)*(2 + x) = 0
$$\left(x - 3\right) \left(x + 2\right) = 0$$
(x - 3)*(x + 2) = 0
Detail solution
Given line equation of 2-order:
$$\left(x - 3\right) \left(x + 2\right) = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
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
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$\left(x - 3\right) \left(x + 2\right) = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
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:
$$\left(x - 3\right) \left(x + 2\right) = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -6$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}1 & 0 & - \frac{1}{2}\\0 & 0 & 0\\- \frac{1}{2} & 0 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = - \frac{25}{4}$$
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
$$\tilde y^{2} - \frac{25}{4} = 0$$
- reduced to canonical form
$$\left(x - 3\right) \left(x + 2\right) = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -6$$
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
$$I_{1} = 1$$
|1 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}1 & 0 & - \frac{1}{2}\\0 & 0 & 0\\- \frac{1}{2} & 0 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
| 1 -1/2| |0 0 |
K2 = | | + | |
|-1/2 -6 | |0 -6|$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = - \frac{25}{4}$$
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
$$\tilde y^{2} - \frac{25}{4} = 0$$
None
- reduced to canonical form