How to solve bivariate quadratic equation

How to solve bivariate quadratic equation

There are two kinds of bivariate quadratic equations: the first is composed of a bivariate quadratic equation and a univariate quadratic equation. It can be solved by directly eliminating the element into a univariate quadratic equation. The second is composed of two bivariate quadratic equations. If it is a common exercise, one (or two) of the equations can be decomposed into two bivariate quadratic equations

How to solve bivariate quadratic equation

There is no formula for binary quadratic equations, so we can only use different methods according to different questions
The first type is a system of equations composed of a bivariate quadratic equation and a bivariate quadratic equation,
a1x+b1y+c1=0 (1)
a2x^2+b2xy+c2y^2+d2x+e2y+f2=0 (2)
It can be transformed into quadratic equation of one variable by substituting elimination method
The second type: a system of equations consisting of two quadratic equations of two variables
a1x^2+b1xy+c1y^2+d1x+e1y+f1=0
a2x^2+b2xy+c2y^2+d2x+e2y+f2=0
(1) If the left side of a bivariate quadratic equation can be factorized, the equation will be factorized into two bivariate quadratic equations, and then form two first type equations with another equation, and then substitute them into elimination. This form of equations generally has four groups of solutions
(2) If it is a system of equations composed of a quadratic equation with one variable and a quadratic equation with two variables, then the quadratic equation with one variable can be solved first, and then it can be substituted into another equation to solve. This kind of system of equations generally has four groups of solutions
(3) If a 1: a 2 = B 1: B 2 = C 1: C 2, the quadratic term can be eliminated and the first type can be solved
(4) If A1: A2 = B1: B2 = D1: D2 or B1: B2 = C1: C2 = E1: E2, the elimination method can be changed to the second form
I wonder if my explanation can help you

How to draw the triangles of equilateral triangle with Geometer's Sketchpad

You can use the value of the point to draw the point in the class. Right click the edge, "draw the point on the line segment", and enter "1 / 3" and "2 / 3" respectively
You can also mark one end point of a line segment as the center, select the other end point, and "transform" - "scale", 1:3, and scale again, with the scale of 2:3

(1) When k takes what value, the solution set of the inequality 2kx ^ 2 + kx-3 / 8mx is {x | 0

1. The inequality 2kx ^ 2 + kx-3 / 8 & lt; 0 holds for all real numbers. If k = 0, substitute it into - 3 / 8 & lt; 0, and holds for all real numbers. If K ≠ 0, then the quadratic function y = 2kx ^ 2 + kx-3 / 8 opens downward and has no intersection with the X axis

In a parallelogram ABCD, where p is on BC and PQ ‖ BD intersects CD with Q, how many triangles are equal to the area of △ ABP in the graph?
BD is diagonal, a in the upper left corner, B in the lower left corner, C in the lower right corner

The parallelogram ABPE is obtained by making PE / / AB through P, and connecting AP, then △ AEP and △ ABP are equal products
Connect be, △ BPE, △ AEB and △ ABP;
If PD is connected, then △ DBP and △ ABP are equal products
So there are four triangles equal to the area of △ ABP
Delta AEP, delta BPE, delta AEB and delta DBP

Given the quadratic function y = - X & # 178; + X-1 / 5, when the independent variable is m, the corresponding function value is greater than 0,
When the independent variables X are M-1 and M + 1, the corresponding function values Y1 and Y2, then Y1 and Y2 must satisfy_____

Given the quadratic function y = - X & # 178; + X-1 / 5, when the independent variable is m, the corresponding function value is greater than 0, when the independent variable is M-1 and M + 1, the corresponding function values Y1 and Y2, then Y1 and Y2 must satisfy the following conditions_ y1+y2＞0____ (Y1 + Y2) / 2 = - M & # 178; + M-1 / 5 ﹥ 0; ﹥ Y1 + Y2 ﹥ 0; glad to answer for you

Put two pieces a and B in the 4-by-4 square diagram. The two pieces are not in the same row or column. How many ways are there? Please provide the process of solving the problem

The first chess piece can be put in a 4 × 4 square. There are 16 different ways to put it
The second piece can be placed in 9 different ways, 3 × 3 = 9, except that the first piece can not be placed in the horizontal and vertical columns
There are 144 different ways to put them

What is the image of the function x squared -- y squared = 1

Hyperbola of a = 1, B = 1

If a, B and C are unequal real numbers and satisfy the relation B2 + C2 = 2A2 + 16A + 14 and BC = A2 -- 4A -- 5, then the value range of a is?
B2 = the square of B
A2 = the square of a
C2 = the square of C
2A2 = the square of 2 * a

Don't understand! @ @ $%^***

As shown in the figure, in the parallelogram ABCD, the bisector of ∠ ABC intersects CD at point E, and the bisector of ∠ ADC intersects AB at point F. try to judge whether AF and CE are equal, and explain the reason

AF = CE. The reasons are as follows: ∵ quadrilateral ABCD is a parallelogram, ∵ ad = CB, ∵ a = C, ∵ ADC = ABC, and ∵ ADF = 12 ≌ ADC, ∵ CBE = 12 ≌ ABC, ∵ ADF = CBE. In △ ADF and △ CBE,