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

__Examples of Systems__

of Quadratic Equations

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

*, another is*

**x**_{1}=3*.*

**x**_{2}=−3*Case*

*:*

**x**_{1}=3

**y**_{1}= −4

**z**_{1}= −5*Case*

*:*

**x**_{2}=−3

**y**_{2}= 4

**z**_{2}= 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

**z**_{1}= −60/12 = −5

**y**_{1}= −60/15 = −4

**x**_{1}= 60/20 = 3*Case*

**x·y·z = −60**Divide it by each given equation, getting

**z**_{2}= 60/12 = 5

**y**_{2}= 60/15 = 4

**x**_{2}= −60/20 = −3*Answer*

There are two solutions to this system:

**(x**_{1},y_{1},z_{1})=(3,−4,−5)

**(x**_{2},y_{2},z_{2})=(−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,

*in terms of*

**y***and substituting into another might be too complex.*

**x**Notice the symmetry between unknown variables

*and*

**x***.*

**y**If, instead of

*, we use*

**x***and, instead of*

**y***, use*

**y***, we will have the same equation.*

**x**In cases like this it might be useful to introduce new variables

*and*

**a=x+y***.*

**b=x·y**Transform these equations:

**(x+2)·(y+2) =**

= x·y+2(x+y)+4 = −10= x·y+2(x+y)+4 = −10

**(x+y)·(x·y−3) =**

= (x+y)·(x·y)−3(x+y) = 15= (x+y)·(x·y)−3(x+y) = 15

We can assign two new unknown variables

*and*

**a=x+y***to obtain the following quadratic system of two equation with one of them being of the first degree*

**b=x·y**

**b + 2a + 4 = −10**

**a·b − 3a = 15**From the first linear equation we can get

*in terms of*

**b***and substitute into a second equation getting a quadratic equation with one variable:*

**a**

**b = −2a − 14**

**a·(−2a−14) − 3a = 15**Simplifying the above:

**2a² + 17a + 15 = 0**This quadratic equation has roots

**a**_{1}= −1

**a**_{2}= −15/2Each of these solutions should be considered separately to get

*and*

**x***.*

**y***Case*

*.*

**a = −1**

**b = −2a−14 = −12**This leads us to a quadratic system

**x + y = −1**

**x·y = −12**Expressing

*in terms of*

**y***from the first equations and substituting into the second:*

**x**

**y = −1 − x**

**x·(−1− x) = −12**

**x² + x − 12 = 0**

**x**_{1}= −4; x_{2}= 3

**y**_{1}= 3; y_{2}= −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

*in terms of*

**y***from the first equations and substituting into the second:*

**x**

**y = −15/2 − x**

**x·(−15/2 − x) = 1**

**x² + (15/2)·x + 1 = 0**

**2x² + 15x + 2 = 0**

**x**

x_{3}= (−15+√209)/4;x

_{4}= (−15−√209)/4

**y**

y_{3}= (−15−√209)/4;y

_{4}= (−15+√209)/4(notice the symmetry, mentioned in the beginning)

*Answer*

There are 4

*pairs of solutions:*

**(x,y)**

**−4, 3**

3, −4

(−15+√209)/4, (−15−√209)/4

(−15−√209)/4, (−15+√209)/43, −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*

*, that is*

**y−1=0**

**y=1**From the second equation of the system

**x·y+y=2**we obtain

*for*

**x***:*

**y=1**

**x·1 + 1 = 2**

**x = 1**So, the first solution to our system is

**(x**_{1}=1, y_{1}=1)*Case*

**x·y+x−2=0**Let's subtract this from the second equation

**x·y+y=2**The result is

*, that is*

**y − x = 0**

**y = x**which converts the second equation into a quadratic one

**x² + x − 2 = 0**Its roots are:

**x**_{2}= −2from which

*and*

**y**_{2}=x_{2}=−2

**x**_{3}= 1from which

**y**_{3}=x_{3}=1which we already obtained in the previous case.

*Answer*

**(x**_{1}=1, y_{1}=1)

**(x**_{2}=−2, y_{2}=−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

*as a function of*

**q***and substitute it the second equation:*

**p**

**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:

*(easy to guess) and*

**p**_{1}=1*(because the product of these roots must be equal to a free member of the equation).*

**p**_{2}=−13/18Correspondingly,

**q**_{1}= 5/36

**q**_{2}= −(13/18)·(5/36) = −65/648Since we have defined

*and*

**p=u+v***, values*

**q=u·v***and*

**u***must be the roots of the quadratic equation*

**v**

**w² −p·w + q = 0***Case*

**(p,q)=(1,5/36)**

**w² − w + 5/36 = 0**Roots are:

*,*

**w**_{1}=1/6

**w**_{2}=5/6Therefore, we have two cases:

*and*

**(u,v)=(5/6, 1/6)**

**(u,v)=(1/6, 5/6)**In the first case we can have a system of equations for

*and*

**x***as*

**y**

**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

*and*

**x***as*

**y**

**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,

*and*

**w**_{1}*will give values for*

**w**_{2}*and*

**u***:*

**v***and*

**(u,v)=(w**_{1}, w_{2})

**(u,v)=(w**_{2}, w_{1})From these two pairs of

*and*

**u***we will derive*

**u***and*

**x***.*

**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.