Lesson Solving systems of algebraic equations of degree 2

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Solving systems of algebraic equations of degree 2

In this lesson we consider the solution of systems of two polynomial equations of degree 2 in two unknowns.

The most general form of this kind of systems is
ax%5E2+%2B+bxy+%2B+cy%5E2+%2B+dx+%2B+ey+%2B+f+=+0

ax%5E2+%2B+bxy+%2B+cy%5E2+%2B+dx+%2B+ey+%2B+f+=+0

, (1)
gx%5E2+%2B+hxy+%2B+iy%5E2+%2B+jx+%2B+ky+%2B+l+=+0

, (2)
where coefficients a, b, c, d , e, f, g, h, i, j, k and l are given numbers.

The method which is used to solve this kind of systems is the "elimination and substitution" method.
The methodology is like this:

First simplify your equations (by performing multiplications of both sides to some numbers, addition and subtraction of equations) to eliminate some terms if required.
Then express one variable (x or y) via another variable from one equation and substitute it to another equation. Finally, get an equation containing only one variable x or y.
When you get such an equation, try to solve it.

If you get the quadratic or bi-quadratic equation, you succeed, because you have formulas to solve them.
But in this way you may also obtain the resultant equation of degree 3 or degree 4 which is not bi-quadratic.
Since the school algebra has no formulas to solve the general algebraic equation of degree 3 or 4, we consider below only those systems of two polynomial equations of degree 2 that may be reduced to quadratic or bi-quadratic resultant equations.

Please note that methods for solution of quadratic equations are described in this site in the lessons
    - PROOF of quadratic formula by completing the square,
    - Introduction into Quadratic Equations and
    - Solving quadratic equations without quadratic formula.

Methods for solution of bi-quadratic equations are described in the lesson Solving algebraic equations of high degree in this site.

As the first form of the system of two polynomial equations of degree 2 we consider equations
ax%5E2+%2B+bxy+%2B+cy%5E2+%2B+f+=+0

, (3)
gx%5E2+%2B+hxy+%2B+iy%5E2+%2B+l+=+0

. (4)
Comparing to the general equations (1), (2), the last two have no terms with x and y of degree 1.

The procedure of solution is as follows:
1) if one of two equations doesn't contain the term x%5E2

(or the term y%5E2

) then apply the substitution method by expressing x (or y) from this equation;
2) if both equations contain the terms x%5E2

and

then perform addition or subtraction of equations first to get an equation, which doesn't contain the term ax%5E2

or the term cy%5E2

. Then apply the substitution method as described in item 1);
3) after applying the substitution method, you will get bi-quadratic equation in one variable;
4) solve the obtained bi-quadratic equation and get the value of the selected variable;
5) substitute the found value of the selected variable to one of the two original equations and get the value of another variable.
An example below help you understand how this procedure works.

Example 1

Solve the system of two equations
x%5E2%2Bxy%2B2y%5E2=74

, (5)
2x%5E2%2B2xy%2By%5E2=73

. (6)

Both equations contain the terms with x%5E2

and

. Therefore, we will first multiply equation (6) by 2 and then subtract equation (5) to eliminate y%5E2

.
By doing so, we will get an equation
3x%5E2%2B3xy=72

.
From the last equation
y=%2824-x%5E2%29%2Fx

. (7)
Substitute this expression to equation (5):
x%5E2+%2Bx%28%2824-x%5E2%29%2Fx%29%2B2%2824-x%5E2%29%5E2%2Fx%5E2=74

.
After simplifying and collecting the common terms you will get
x%5E4-73x%5E2%2B576=0

.
This is bi-quadratic equation. Introducing z=x%5E2

, we reduce it to the quadratic equation
z%5E2-73z%2B576=0

.
Apply the quadratic formula. You obtain
z+=+%2873%2B-sqrt%2873%5E2-4%2A1%2A576%29%29%2F2

.
First value of z gives two solutions for x: x%5B1%5D=8

.
Second value of z gives two other solutions for x: x%5B3%5D=3

.
Substituting these four values of x into expression (5) gives four solutions for y:
y%5B1%5D=-5

.

As the second form of the system of two polynomial equations of degree 2 we consider the following

Example 2

Solve the system of two equations
%28x-1%29%5E2%2B%28y-1%29%5E2=4

, (8)
%28x-2%29%5E2%2B%28y-2%29%5E2=2

. (9)

Equation (8) represents the circle of the radius 2 with the center at the point (1,1).
This circle is shown in Figure 1 by the red line.
Equation (9) represents the circle of the radius sqrt%282%29

with the center at the point (2,2).
This circle is shown in Figure 1 by the blue line.

Let us open parentheses and collect the common terms in equations (8) and (9). We will get
x%5E2-2x%2By%5E2-2y=2

, (10)
x%5E2-4x%2By%5E2-4y=-6

. (11)

Subtracting (11) from (10) leads to
2x%2B2y=8

,
or

. (12)

Figure 1. To the Example 2

Substitute this to equation (10) and collect the common terms. It gives a quadratic equation for y:
y%5E2-4y%2B3=0

. (13)

Apply the quadratic formula
y=%284%2B-sqrt%284%5E2-4%2A1%2A3%29%29%2F2+=+%284%2B-sqrt%284%29%29%2F2

to get the roots of quadratic equation (13) y%5B1%5D

and

.
So, two solutions of the system (8), (9) are:
x%5B1%5D=1

and

.
Note that these solutions represent exactly intersection points of two circles in Figure 1.

Note that the system of equations (10), (11) is different in their form from (3), (4) due to presence of terms with x and y of degree one. Nevertheless, the system (10), (11) still can be solved by the "elimination and substitution" method. The success in this example is due to the fact that elimination leads to the linear equation in this case. In turn, this fact is the direct consequence of the proportionality of coefficients at the higher degree terms x%5E2

and

in equations (10), (11).

Third form of the system of two polynomial equations of degree 2 we illustrate with the following

Example 3

Solve the system of two equations
y=%28x-1%29%5E2-2

, (14)
y=0.5%28%282x-3%29%5E2-5%29

. (15)

Equation (14) represents the parabola with the vertex at x=1.
This parabola is shown in Figure 2 by the red line.
Equation (15) represents the parabola with the vertex at x=1.5.
This parabola is shown in Figure 2 in green color.

Let us open parentheses, multiply equation (14) by 2 and collect the common terms. We will get
x%5E2-2x-y-1=0

, (16)
4x%5E2-12x-2y-5=0

. (17)

Figure 2. To the Example 3

The distinctive feature of equations (16), (17) comparing with (1), (2), is the absence of the terms y%5E2

and

.

Multiply equation (16) by 2 and subtract from (17) to eliminate variable y. You will get one quadratic equation for variable x
2x%5E2-8x-6=0

, or

. (18)
The last equation has the roots x%5B1%5D=3

and

(use the quadratic formula or Viete's formulas).
Substituting these values of x to equation (12), you will get two solutions of the system (14), (15):
x%5B1%5D=3

and

.

Note that these solutions represent two intersection points of parabolas in Figure 2.

Also note that you could get the same equation (18) and the same solutions by equating right sides of equations (14) and (15). It is simply another way to make elimination.

As the fourth form of the system of two polynomial equations of degree 2 we consider the following

Example 4

Solve the system of two equations
x%5E2%2By%5E2=4

, (19)
xy=1

. (20)

Equation (19) represents the circle of the radius 2 with the center point x=0

.
This circle is shown in Figure 3 by the red and green lines.
Equation (20) represents the hyperbola with the center point x=0

.
This hyperbola is shown in Figure 3 in blue color.

Using the elimination method we have from (20)
y=1%2Fx

.

Substituting this to equation (19) leads to
x%5E2+%2B+1%2Fx%5E2+=+4

Figure 3. To the Example 4

Solving this bi-quadratic equation gives the roots
x%5E2=+%284%2B-sqrt%284%5E2-4%2A1%2A1%29%29%2F2+=+2%2B-sqrt%283%29

, or

.

So, four solutions of the system (19), (20) are:
x%5B1%5D=sqrt%282%2Bsqrt%283%29%29

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