Question

1. (x + 2)(x − 5) < 0
2. x 2 > 16
3. x 2 − 9 < 0
4. x(x + 1) > 0
5. x 2 > 4x
6. x 2 + 2x + 1 > 0
7. x 2 − x − 6 ≤ 0
8. 3x 2 < 10 − x
9. x 2 − 2x − 5 ≥ 3
10. x 2 < 5
11. 6x − 8 > x 2
12. 6x 2 + x − 12 < 0
13. x(3x − 1) ≤ 4
14. 25x 2 − 9 ≤ 0
15. 2x 2 < 5x + 3

Answer 1

Se trata de resolver las Inecuaciones Cuadráticas suministradas.

1) (x + 2)(x – 5) < 0

x² + 2x – 5x – 10 < 0

x² – 3x – 10 < 0

Se soluciona mediante la Resolvente de la Ecuación de Segundo Grado.

X₁,₂ = – B ± √(B² – 4AC) ÷ 2A

A: coeficiente que acompaña al termino cuadrático.

B: coeficiente que acompaña al termino elevado a la unidad.

C: Coeficiente del término independiente o constante.

Para este caso se tiene:

A = 1; B = – 3; C = – 10

X1,2 = – (– 3) ± √[(– 3)² – 4(1)(– 10)] ÷ 2(1)

X1,2 = 3 ± √(9 +40) ÷ 2

X1,2 = 3 ± √49 ÷ 2

X1,2 = 3 ± 7 ÷ 2

X1 = 3 + 7 ÷ 2

X1 = 10 ÷ 2

X1 = 5

X2 = 3 – 7 ÷ 2

X2 = – 4 ÷ 2

X2 = – 2

2) x² > 16

x > √16

x > 4

3) x² – 9 < 0

x² < 9

x < √9

x < 3

4) x(x + 1) > 0

x² + x > 0

Usando la Resolvente.

A = 1; B = 1; C = 0

X1,2 = – (1) ± √[(1)² – 4(1)(0)] ÷ 2(1)

X1,2 = – 1 ± √1 ÷ 2

X1,2 = – 1 ± 1 ÷ 2

X1 = – 1 + 1 ÷ 2

X1 = 0

X2 = – 1 – 1 ÷ 2

X2 = – 2 ÷ 2

X2 = – 1

5) x² > 4x

x2 – 4x > 0

A = 1; B = – 4; C = 0

X1,2 = – (– 4) ± √[(– 4)² – 4(1)(0)] ÷ 2(1)

X1,2 = 4 ± √(16) ÷ 2

X1,2 = 4 ± 4 ÷ 2

X1 = 4 + 4 ÷ 2

X1 = 8 ÷ 2

X1 = 4

X2 = 4 – 4 ÷ 2

X2 = 0

6) x² + 2x + 1 > 0

A = 1; B = 2; C = 1

X1,2 = – (2) ± √[(2)² – 4(1)(1)] ÷ 2(1)

X1,2 = – 2 ± √[(4 – 4] ÷ 2

X1,2 = – 2 ± 0 ÷ 2

X1,2 = – 2 ÷ 2

X1,2 = – 1

7) x² – x – 6 ≤ 0

A = 1; B = – 1 ; C = – 6

X1,2 = – (– 1) ± √[(– 1)² – 4(1)(– 6)] ÷ 2(1)

X1,2 = – 1 ± √(1 + 24) ÷ 2

X1,2 = – 1 ± √(25) ÷ 2

X12 = – 1 ± 5 ÷ 2

X1 = – 1 + 5 ÷ 2

X1 = 4 ÷ 2

X1 = 2

X2 = – 1 – 5 ÷ 2

X2 = – 6 ÷ 2

X2 = – 3

8) 3x² < 10 – x

3x2 + x – 10 < 0

A = 3; B = 1; C = – 10

X1,2 = – (1) ± √[(1)² – 4(3)(– 10)] ÷ 2(3)

X1,2 = – 1 ± √(1 + 120) ÷ 6

X1,2 = – 1 ± √(121) ÷ 6

X1,2 = – 1 ± 11 ÷ 6

X1 = – 1 + 11 ÷ 6

X1 = 10 ÷ 6

X1 = 5/3

X2 = – 1 – 11 ÷ 6

X2 = – 12 ÷ 6

X2 = – 2

9) x² – 2x – 5 ≥ 3

x² – 2x – 5 – 3≥ 0

x² – 2x – 8 ≥ 0

A = 1; B = – 2 = C = – 8

X1,2 = – (– 2) ± √[(– 2)² – 4(1)(– 8)] ÷ 2(1)

X1,2 = 2 ± √(4 + 32) ÷ 2

X1,2 = 2 ± √(36) ÷ 2

X1,2 = 2 ± 6 ÷ 2

X1 = 2 + 6 ÷ 2

X1 = 8 ÷ 2

X1 = 4

X2 = 2 – 6 ÷ 2

X2 = – 4 ÷ 2

X2 = – 2

10) x² < 5

x² – 5 < 0

A = 1; B = 0; C = – 5

X1,2 = – (0) ± √[(0)² – 4(1)(– 5)] ÷ 2(1)

X1,2 = ± √(20) ÷ 2

X1,2 = ± 4,47 ÷ 2

X1,2 = ± 2,236

11) 6x – 8 > x²

0 > x² – 6x + 8

A = 1; B = – 6; C = 8

X1,2 = – (– 6) ± √[(– 6)² – 4(1)(8)] ÷ 2(1)

X1,2 = 6 ± √(36 – 32) ÷ 2

X1,2 = 6 ± √(4) ÷ 2

X1,2 = 6 ± 2 ÷ 2

X1 = 6 + 2 ÷ 2

X1 = 8 ÷ 2

X1 = 4

X2 = 6 – 2 ÷ 2

X2 = 4 ÷ 2

X2 = 2

12) 6x² + x – 12 < 0

A = 6; B = 1; C = – 12

X1,2 = – (1) ± √[(1)² – 4(6)(– 12)] ÷ 2(6)

X1,2 = – 1 ± √(1 + 288) ÷ 12

X1,2 = – 1 ± 17 ÷ 12

X1 = – 1 + 17 ÷ 12

X1 = 16 ÷ 12

X1 = 4/3

X2 = – 1 – 17 ÷ 12

X2 = – 18 ÷ 12

X2 = – 3/2

13) x(3x – 1) ≤ 4

x(3x – 1) ≤ 4

3x² – x ≤ 4

3x² – x – 4 ≤ 0

A = 3; B = – 1; C = – 4

X1,2 = – (– 1) ± √[(– 1)² – 4(3)(– 4)] ÷ 2(3)

X1,2 = 1 ± √(1 + 48) ÷ 6

X1,2 = 1 ± √(49) ÷ 6

X1,2 = 1 ± 7 ÷ 6

X1 = 1 + 7 ÷ 6

X1 = 8 ÷ 6

X1 = 4/3

X2 = 1 – 7 ÷ 6

X2 = – 6 ÷ 6

X2 = – 1

14) 25x² – 9 ≤ 0

A = 25; B = 0; C = – 9

X1,2 = – (0) ± √[(0)² – 4(25)(– 9)] ÷ 2(25)

X1,2 = ± √900 ÷ 50

X1,2 = ± 30 ÷ 50

X1,2 = ± 3/5

15) 2x² < 5x + 3

2x2 – 5x – 3 < 0

A = 2; B = – 5; C = – 3

X1,2 = – (– 5) ± √[(– 5)² – 4(2)(– 3)] ÷ 2(2)

X1,2 = 5 ± √(25 + 24) ÷ 4

X1,2 = 5 ± √49 ÷ 4

X1,2 = 5 ± 7 ÷ 4

X1 = 5 + 7 ÷ 4

X1 = 12 ÷ 4

X1 = 3

X2 = 5 – 7 ÷ 4

X2 = – 2 ÷ 4

X2 = – 1/2