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.
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.
Los ceros de una función son los valores donde la misma se hace cero.
Encontramos los ceros : 1.
X ^ 2– 3x = 0x * (x - 3) = 0 Entonces x = 0 ó x = 32.
6x ^ 2 + 42x = 06x * (x + 7) = 0 Entonces x = 0 ó x = - 73.
X ^ 2 + 8x = 0x * (x + 8) = 0 Entonces x = 0 ó x = - 84.
X(2x – 3) – 3(5 – x) = 832x ^ 2 - 3x - 15 + 3x - 83 = 02x ^ 2 - 98 = 02x ^ 2 = 98x ^ 2 = 98x = raiz(68) ó x = - raiz(68)5.
(2x + 5)(2x – 5) = 112 * (x + 2.
5) * 2 * (x - 2.
5) = 0Entonces x = 2.
5 ó x = - 2.
58. (7 + x) ^ 2 + (7 – x) ^ 2 = 13049 + 14x + x ^ 2 + 49 - 14x + x ^ 2 = 1302x ^ 2 + 2 * 49 = 130x ^ 2 + 49 = 65x ^ 2 = 65 - 49 = 16x = 4 ó x = - 4.
x ^ 6 = 64 x = ( + - ) raíz sexta de 64 x1 = 2 x2 = - 2.
La solución a tu problema es la siguiente 6 - 6 = 0.