Question

Encontrar las raíces o ceros de las siguientes funciones utilizando cualquiera de los dos métodos estudiados.
1. x2 – 3x = 0
2. 6x2 + 42x = 0
3. x2 + 8x = 0
4. x(2x – 3) – 3(5 – x) = 83
5. (2x + 5)(2x – 5) = 11
6. x/x+1+x+1/x=13/6
7. 4/x-1-3-x/2=2
8. (7 + x)2 + (7 – x)2 = 130
9. 8(2 – x)2 = 2(8 – x)2
10. 3x+54/2x+3=18

Answer 1

Respuesta:

1) x1 = 3; x2 = 0

2) x1 = 3; x2 = 0

3) x1 = 0; x2 = - 8

4) x1 = 7; x2 = - 7

5) x1 = 3; x2 = - 3

6) x1 = 0,245; x2 = 24,755

7) x1 = -1,24; x2 = 7,24

8) x1 = 4; x2 = - 4

9) x1 = 4; x2 = - 4

10) x = 0

Respuesta Paso a Paso:

1) x² – 3x = 0

Se aplica la Ecuación de Segundo Grado, donde los términos son:

A = 1; B = -3; C = 0

La fórmula general de la ecuación cuadrática es:

x(1,2) = - B ± √(B² – 4A x C) ÷ 2A

Aplicando:

x(1,2) = - (-3) ± √[(- 3)2 – 4(1)(0)] ÷ 2(1)

x(1,2) = 3 ± √(9) ÷ 2 = 3 ± 3 ÷ 2

x1 = 3 + 3 ÷ 2 = 3 + 3 ÷ 2 = 6 ÷ 2 = 3

x1 = 3

x2= 3 - 3 ÷ 2 = 0

x2 = 0

2) 6x2 + 42x = 0

A = 6; B = 42; C = 0

x(1,2) = - (42) ± √[(42)2 – 4(6)(0)] ÷ 2(6)

x(1,2) = 3 ± √(9) ÷ 2 = 3 ± 3 ÷ 2

x1 = 3 + 3 ÷ 2 = 3 + 3 ÷ 2 = 6 ÷ 2 = 3

x1 = 3

x2= 3 - 3 ÷ 2 = 0

x2 = 0

3) x² + 8x = 0

A = 1; B = 8; C = 0

x(1,2) = - (8) ± √[(8)² – 4(1)(0)] ÷ 2(1)

x(1,2) = - 8 ± √(64) ÷ 2 = - 8 ± 8 ÷ 2

x1 = - 8 + 8 ÷ 2 =0

x1 = 0

x2= - 8 - 8 ÷ 2 = - 16 ÷ 2 = - 8

x2 = - 8

4) x(2x – 3) – 3(5 – x) = 83

Resolviendo los binomios queda:

2x² – 3x – 15 + 3x = 83

2x² - 98 = 0  

A = 2; B = 0; C = - 98

x(1,2) = - (0) ± √[(0)² – 4(2)(- 98)] ÷ 2(2)

x(1,2) = ± √(784) ÷ 4 = -± 28 ÷ 4

x1 = 28 ÷ 4 = 7

x1 = 7

x2= -28 ÷ 4 = - 28 ÷ 4 = - 7

x2 = - 7

5) (2x + 5)(2x – 5) = 11

Resolviendo los binomios queda:

4x² - 10x + 10x – 25 = 11

4x² – 25 – 11 = 0

4x² – 36 = 0

A = 4; B = 0; C = - 36

x(1,2) = - (0) ± √[(0)² – 4(4)(- 36)] ÷ 2(4)

x(1,2) = ± √(576) ÷ 8 = -± 24 ÷ 8

x1 = 24 ÷ 8 = 3

x1 = 3

x2= - 24 ÷ 8 = - 24 ÷ 8 = - 3

x2 = - 3

6) x/(x+1) + (x+1)/x = 13/6

Al resolver la expresión queda:

- x² + 25x + 6 = 0

A = - 1; B = 25; C = 6

x(1,2) = - (25) ± √[(25)² – 4(- 1)(6)] ÷ 2(- 1)

x(1,2) = - 25 ± √(625 + 24) ÷- 2 = - 25 ± √(601) ÷- 2 = - 25 ± 24,51 ÷- 2

x1 = - 25 + 24,51 ÷- 2 = - 0,49 ÷- 2 = 0,245

x1 = 0,245

x2= - 25 - 24,51 ÷- 2 = - 49,51 ÷- 2 = 24,755

x2 = 24,755

7) 4/x – 1 – 3 - x/2 = 2

Al resolver la expresión queda:

- x² - 6x + 9 = 0

A = -1; B = -6; C = 9

x(1,2) = - (- 6) ± √[(- 6)² – 4(- 1)(9)] ÷ 2(- 1)

x(1,2) = - 6 ± √[(36) + 36] ÷ - 2 = -6 ± √72 ÷ - 2 = -6 ± 8,48 ÷ - 2

x1 = - 6 + 8,48 ÷ - 2 = 2,48 ÷ - 2 = - 1,24

x1 = - 1,24

x2= - 6 - 8,48 ÷ - 2 = - 14,48 ÷ - 2 = 7,24

x2 = 7,24

8) (7 + x)² + (7 – x)² = 130

Resolviendo los binomios cuadrados queda:

2x² – 32 = 0

A = 2; B = 0; C = - 32

x(1,2) = - (0) ± √[(0)² – 4(2)(- 32)] ÷ 2(2)

x(1,2) = ± √(256) ÷ 8 = -± 16 ÷ 4

x1 = 16 ÷ 4 = 4

x1 = 4

x2= - 16 ÷ 4 = - 16 ÷ 4 = - 4

x2 = - 4

9) 8(2 – x)² = 2(8 – x)²

Resolviendo los binomios cuadrados queda:

6x² -96 = 0

A = 6; B = 0; C = - 96

x(1,2) = - (0) ± √[(0)² – 4(6)(- 96)] ÷ 2(6)

x(1,2) = ± √(2.304) ÷ 12 = -± 48 ÷ 12

x1 = 48 ÷ 12 = 4

x1 = 4

x2= - 48 ÷ 4 = - 4

x2 = - 4

10) 3x + 54/2x + 3 = 18

Resolviendo la expresión queda:

-33x = 0

x = 0

Solamente tiene una raíz y negativa.

Answer 2

Los ceros de una función son los valores donde la misma se hace cero.

Encontramos los ceros:

1. x^2– 3x = 0

x*(x-3) =0 Entonces x = 0 ó x = 3

2. 6x^2 + 42x = 0

6x*(x + 7) = 0 Entonces x = 0 ó x = -7

3. x^2 + 8x = 0

x*(x+8) = 0 Entonces x = 0 ó x = -8

4. x(2x – 3) – 3(5 – x) = 83

2x^2 -3x -15+ 3x -83 = 0

2x^2 - 98 = 0

2x^2 = 98

x^2 = 98

x = raiz(68) ó x = -raiz(68)

5. (2x + 5)(2x – 5) = 11

2*(x + 2.5)*2*(x- 2.5) = 0

Entonces x = 2.5 ó x = -2.5

8. (7 + x)^2 + (7 – x)^2= 130

49 +14x + x^2 +49 -14x + x^2 = 130

2x^2 + 2*49 = 130

x^2 + 49 = 65

x^2 = 65-49 = 16

x = 4 ó x = -4