Systems of Quadratic Equations: UNIZOR.COM - Math4Teens - Algebra - Equa...

Notes to a video lecture on http://www.unizor.com

Examples of Systems
of Quadratic Equations

We will restrict solutions to real numbers only.

Example 1
x·y = −12
x·z = −15
y·z = 20

Solution A
From the third equation:
z = 20/y
Substitute it to the second, getting a system of two equations with two unknowns:
x·y = −12
x·20/y = −15
From the second equation:
y = −x·20/15 = −x·4/3
Substitute into the first equation:
−x·x·4/3 = −12
Solving it for x:
x² = 9
One root is x₁=3, another is x₂=−3.
Case x₁=3:
y₁ = −4
z₁ = −5
Case x₂=−3:
y₂ = 4
z₂ = 5

Solution B (more clever)
Obviously, none of the unknowns equals to zero.
Multiply all three equations, getting
(x·y·z)² = 3600
Therefore, we have two cases:
x·y·z = 60
or
x·y·z = −60
Case x·y·z = 60
Divide it by each given equation, getting
z₁ = −60/12 = −5
y₁ = −60/15 = −4
x₁ = 60/20 = 3
Case x·y·z = −60
Divide it by each given equation, getting
z₂ = 60/12 = 5
y₂ = 60/15 = 4
x₂ = −60/20 = −3

Answer
There are two solutions to this system:
(x₁,y₁,z₁)=(3,−4,−5)
(x₂,y₂,z₂)=(−3,4,5)

Checking
ALWAYS PERFORM CHECKING OF ALL SOLUTIONS


Example 2
(x+2)·(y+2) = −10
(x+y)·(x·y−3) = 15

Solution
Both equations are of the second degree, so directly resolving one of them for, say, y in terms of x and substituting into another might be too complex.
Notice the symmetry between unknown variables x and y.
If, instead of x, we use y and, instead of y, use x, we will have the same equation.
In cases like this it might be useful to introduce new variables a=x+y and b=x·y.
Transform these equations:
(x+2)·(y+2) =
= x·y+2(x+y)+4 = −10
(x+y)·(x·y−3) =
= (x+y)·(x·y)−3(x+y) = 15
We can assign two new unknown variables a=x+y and b=x·y to obtain the following quadratic system of two equation with one of them being of the first degree
b + 2a + 4 = −10
a·b − 3a = 15
From the first linear equation we can get b in terms of a and substitute into a second equation getting a quadratic equation with one variable:
b = −2a − 14
a·(−2a−14) − 3a = 15
Simplifying the above:
2a² + 17a + 15 = 0
This quadratic equation has roots
a₁ = −1
a₂ = −15/2
Each of these solutions should be considered separately to get x and y.

Case a = −1.
b = −2a−14 = −12
This leads us to a quadratic system
x + y = −1
x·y = −12
Expressing y in terms of x from the first equations and substituting into the second:
y = −1 − x
x·(−1− x) = −12
x² + x − 12 = 0
x₁ = −4; x₂ = 3
y₁ = 3; y₂ = −4
(notice the symmetry, mentioned in the beginning)

Case a = −15/2.
b = −2a−14 = 1
This leads us to a quadratic system
x + y = −15/2
x·y = 1
Expressing y in terms of x from the first equations and substituting into the second:
y = −15/2 − x
x·(−15/2 − x) = 1
x² + (15/2)·x + 1 = 0
2x² + 15x + 2 = 0
x₃ = (−15+√209)/4;
x₄ = (−15−√209)/4
y₃ = (−15−√209)/4;
y₄ = (−15+√209)/4
(notice the symmetry, mentioned in the beginning)

Answer
There are 4 (x,y) pairs of solutions:
−4, 3
3, −4
(−15+√209)/4, (−15−√209)/4
(−15−√209)/4, (−15+√209)/4

Checking
ALWAYS PERFORM CHECKING OF ALL SOLUTIONS


Example 3
x·y² − x − 2y = −2
x·y + y = 2

Solution
The first equation can be transformed to an equivalent
x·(y²−1) − 2(y−1) = 0
x·(y−1)·(y+1) − 2(y−1) = 0
(y−1)·(x·y+x−2) = 0
This allows to split a problem in two:
Case y−1=0, that is y=1
From the second equation of the system x·y+y=2
we obtain x for y=1:
x·1 + 1 = 2
x = 1
So, the first solution to our system is
(x₁=1, y₁=1)
Case x·y+x−2=0
Let's subtract this from the second equation x·y+y=2
The result is
y − x = 0, that is y = x
which converts the second equation into a quadratic one
x² + x − 2 = 0
Its roots are:
x₂ = −2
from which y₂=x₂=−2 and
x₃ = 1
from which y₃=x₃=1
which we already obtained in the previous case.

Answer
(x₁=1, y₁=1)
(x₂=−2, y₂=−2)

Checking
DON'T FORGET TO CHECK YOUR SOLUTIONS

Example 4
1/(x+y) + 1/(x−y) = 36/5
(x+y)² + (x−y)² = 13/18

Solution
First simplification can be achieved by introducing two new variables:
u = x+y
v = x−y
This leads to a new simpler system of equations:
1/u + 1/v = 36/5
u² + v² = 13/18
The first on can be rewritten as:
(u+v)/(u·v) = 36/5
The second one can be rewritten as:
u² + 2u·v + v² − 2u·v = 13/18
or
(u+v)² − 2u·v = 13/18
Another pair of variables can simplify it even further:
p = u+v
q = u·v
In terms of these variables the system looks like
p/q = 36/5
p² − 2q = 13/18
Now we can express q as a function of p and substitute it the second equation:
q = 5·p/36
p² − 2·5·p/36 − 13/18 = 0
So, we have a plain quadratic equation for p:
p² − (5/18)p − 13/18 = 0
It has 2 roots: p₁=1 (easy to guess) and p₂=−13/18 (because the product of these roots must be equal to a free member of the equation).
Correspondingly,
q₁ = 5/36
q₂ = −(13/18)·(5/36) = −65/648
Since we have defined p=u+v and q=u·v, values u and v must be the roots of the quadratic equation w² −p·w + q = 0
Case (p,q)=(1,5/36)
w² − w + 5/36 = 0
Roots are:
w₁=1/6, w₂=5/6
Therefore, we have two cases:
(u,v)=(5/6, 1/6) and
(u,v)=(1/6, 5/6)
In the first case we can have a system of equations for x and y as
x+y = u = 5/6
x−y = v = 1/6
Solution to this linear system is:
x = 1/2, y = 1/3
In the second case we can have a system of equations for x and y as
x+y = u = 1/6
x−y = v = 5/6
Solution to this linear system is:
x = 1/2, y = −1/3
Case (p,q)=(−13/18,−65/648)
w² + (13/18)·w − 65/648 = 0
Roots of this equation, w₁ and w₂ will give values for u and v:
(u,v)=(w₁, w₂) and
(u,v)=(w₂, w₁)
From these two pairs of u and u we will derive x and y.
Getting exact values in this case we leave to those who are not afraid of cumbersome calculations.

Checking
JUST DON'T FORGET TO CHECK YOUR SOLUTION.