Complemento Teoria y Ejercicios

8/18/2019 Complemento Teoria y Ejercicios

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1. El sentido de una desigualdad no se altera, si a cada miembro se suma o se resta un número.

2. Una desigualdad que es válida para algunos valores reales de la variable, pero no lo es para otros, es u

desigualdad absoluta.

3. Una desigualdad que es válida para todos los valores reales de la variable, es una desigualdad

condicional

4. Una fracción racional es el cociente de dos polinomios.

5. La ecuación más general de segundo grado con dos variables es la ecuación del tipo

ax2+bxy+cy2+dx+ey+f=0

6. Una ecuación es la proposición de que dos expresiones son iguales.

7. Las ecuaciones que son válidas para todos los valores posibles de las letras que contienen se llaman

identidades

8. Las ecuaciones que son válidas para algunos valores de sus letras, pero que no lo son para otros, sellaman ecuaciones condicionales.


2/18

lve for x :

3 x

4

2 x 1

3

11

12

Put the fractions in2 3 x

4

2 x 1

3

over a common denominator.

t each term in

2 3 x

4

2 x 1

3 over the common denominator 12:

2 3

4

x 1

3

3 2 3 x

12

4 2 x 1

12:

3 2 3 x

12

4 2 x 1

12

11

12

Combine3 2 3 x

12

4 2 x 1

12into a single fraction.

2 3 x

12

4 2 x 1

12

3 2 3 x 4 2 x 1

12:

3 2 3 x 4 2 x 1

12

11

12


3/18

olve for x over the real numbers:

3 x 6 4 x 4

Eliminate the square roots on both sides.

quare both sides:

x 6 4 x 4

Isolate x to the left hand side.

ubtract 4 x 6 from both sides:

x

10


4/18

lve for x :

2 x j 2 k 2 j x 2 0

Write the quadratic equation in standard form.

vide both sides by 2 j:

k2

2 j

x j 2 k

2 j x 2 0

Solve the quadratic equation by completing the square.

dk2

2 jto both sides:

j 2 k

2 j x 2

k2

2 j

Take one half of the coefficient of x

and square it, then add it to both sides.

d j 2 k

2

16 j2to both sides:


5/18

2x^2x30

Input:

2 x 2 x 3 0

Inequality plot:

2 1 1 2

2

2

4

6

8

Alternate forms:

2 x 2 x 3

x 1 x 3

2

0

x 1 2 x 3 0

Alternate form assuming x is positive:

x 1

Solution: Approximate form

3

2 x 1

Integer solutions:

x 1

x 0

Number line:

1.5 1.0 0.5 0.0 0.5 1.0

Generated by Wolfram|Alpha (www.wolframalpha.com) on April 24, 2016 from Champaign, IL.

© Wolfram Alpha LLC— A Wolfram Research Company

1


6/18

Interval notation:

3

2, 1

2x^2+x-3


7/18

2x^23y^21, 4x^25y^25

Input:

2 x 2 3 y 2 1, 4 x 2 5 y 2 5

Plot of solution set:

3 2 1 0 1 2 3

4

2

0

2

4

x

y

2 x 23 y 2 1

4 x 25 y 2 5

Alternate form:

2 x 2 3 y 2 1, 4 x 2 5 y 2 1

Solutions: Approximate forms

x 5 , y 3

x 5 , y 3

x 5 , y 3

x 5 , y 3



1


8/18

olve for x :

x 2 3 0

Divide both sides by a constant to simplify the equation.

ivide both sides by 5:

3 0

Isolate terms with x to the left hand side.

ubtract 3 from both sides:

3

Eliminate the exponent on the left hand side.


9/18


x 3 2 x 2

Move everything to the left hand side.

ubtract 2 x 2 from both sides:

2 x 2 6 x 3 0

Write the quadratic equation in standard form.


3 x

3

2

0

Solve the quadratic equation by completing the square.


10/18


x 6 19 x 3 27

Move everything to the left hand side.

ubtract 19 x 3 27 from both sides:

x 6 19 x 3 27 0

Simplify 8 x 6 19 x

3 27 0 by making a substitution.

ubstitute y x 3:

y 2 19 y 27 0

Factor the left hand side.


11/18


x 3 3 x 2 x 4 0

Divide both sides by a constant to simplify the equation.


3 x 2 x 4 0

Solve each term in the product separately.

plit into two equations:

0 or 3 x 2 x 4 0

Look at the first equation:Elminate the power on the left hand side.


12/18

a^16b^160

Input:

a 16

b 16

0

Solutions: More roots Approximate forms

Stepbystep solution

b a

b a

b a

b a

b 18

a

Integer solution:

a 0 , b 0



1


13/18

lve for x :

d d x c x 2 0

Using the quadratic formula, solve for x .

d d2 4 c c d

2 c:

d2 4 c c d d

2 cor x

d d2 4 c c d

2 c

Look at the first equation: Expand polynomials.

4 c c d 4 c2 4 c d d

2:

4 c2 4 c d d

2 d

2 cor x

d d2 4 c c d

2 c

Look at the second equation: Expand polynomials.


14/18

mpute the partial fraction decomposition of the following:

6 x 9

x 5 2 x 1

e partial fraction expansion is of the form:

6 x 9

x 5 2 x 1

Θ 1

2 x 5

Θ 2

2 x 1

ultiply both sides by 2 x 5 2 x 1 and simplify:x 9 Θ 1 2 x 1 Θ 2 2 x 5

pand and collect in terms of powers of x :

x

9Θ

1

5Θ

2

2Θ

1

2Θ

2 x

uate coefficients on both sides, yielding 2 equations in 2 unknow

Θ 1 5 Θ 2

2 Θ 1 2 Θ 2

matrix form the system is written as:

1 5

2 2

Θ 1

Θ 2

9

6


15/18

mpute the partial fraction decomposition of the following:

6 x 11

x 52

e partial fraction expansion is of the form:

6 x 11

x 52

Θ 1

2 x 5

Θ 2

2 x 52

ultiply both sides by 2 x 52 and simplify:

x 11 Θ 2 Θ 1 2 x 5

pand and collect in terms of powers of x :

x 11 5 Θ 1 Θ 2 2 Θ 1 x

uate coefficients on both sides, yielding 2 equations in 2 unknow

1 5 Θ 1 Θ 2

2 Θ 1

matrix form the system is written as:

1

0

Θ 1

Θ 2

11

6


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17/18


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x42

Input:

x 4 2

z is the absolute value of z »

Inequality plot:

5 5 10 15

2

4

6

8

10

12

Alternate form assuming x is real:

x

4

2

2

Solution:

2 Rex 6 , Rex 2

8 Rex 12 Imx Rex 2

8 Rex 12

Re z is the real part of z »

Im z is the imaginary part of z »

Integer solutions:

x 3

x 4

x 5

Number line:

2 3 4 5 6

Generated by Wolfram|Alpha (www wolframalpha com) on April 24 2016 from Champaign IL

Complemento Teoria y Ejercicios

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