ECUACIONES

Álgebra Polinomial  1º BCT

ECUACIONES DE GRADO SUPERIOR A 2

Resolver las siguientes ecuaciones:

3 2
1) x + 3x – x – 3 = 0
3 2
1 3 -1 -3 x + 3x – x – 3 = (x – 1)(x + 1)(x + 3)
1 1 4 3
 x − 1 = 0 ⇒ x1 = 1
1 4 3 0 

(x – 1)(x + 1)(x + 3) = 0 ⇒  x + 1 = 0 ⇒ x 2 = −1
-1 -1 -3 
1 3 0  x + 3 = 0 ⇒ x 3 = −3

3 2
2) x + 2x + 2x + 1 = 0
3 2 2
1 2 2 1 x + 2x + 2x + 1 = (x + 1) · (x + x + 1)
-1 -1 -1 -1  x − 1 = 0 ⇒ x1 = 1
(x – 1) (x2 + x + 1) = 0 ⇒ 
1 1 1 0  x 2 + x + 1 = 0

Las otras dos raíces las calculamos aplicando la fórmula de la ecuación de segundo grado:
−1 ± 1 − 4
x2 + x + 1 = 0 ⇒ x = → No tiene solución
2

3 2
3) x + 3x – 4x – 12 = 0
3 2
1 3 -4 - 12 x + 3x – 4x – 12 = (x – 2)(x + 2)(x + 3)
2 2 10 12
 x − 2 = 0 ⇒ x1 = 2
1 5 6 0 

(x – 2)(x + 2)(x + 3) = 0 ⇒  x + 2 = 0 ⇒ x 2 = −2
-2 -2 -6 
1 3 0  x + 3 = 0 ⇒ x 3 = −3

3 2
4) x – x – x + 1 = 0

1 -1 -1 1 3 2 2
x – x – x + 1 = (x + 1 )(x – 1)
-1 -1 2 -1
1 -2 1 0  x − 1 = 0 ⇒ x1 = 1
2 
(x + 1 )(x – 1) = 0 ⇒ 
1 1 -1  x + 1 = 0 ⇒ x 2 = −1
1 -1 0

3 2
5) x – 2x – 4x + 8 = 0

1 -2 -4 8 3 2 2
x – 2x – 4x + 8 = (x + 2)(x – 2)
2 2 0 -8

 x − 2 = 0 ⇒ x1 = 2
1 0 -4 0
2
(x + 2)(x – 2) = 0 ⇒ 
2 2 4  x + 2 = 0 ⇒ x 2 = − 2
1 2 0

Luisa Muñoz 1
ECUACIONES 1º BCT

3 2
6) 6x + 7x – 9x + 2 = 0
3 2 2
6 7 -9 2 6x + 7x – 9x + 2 = (x + 2)(6x - 5x + 1)
-2 -12 10 -2
2
 x + 2 = 0 ⇒ x1 = −2
6 -5 1 0 (x + 2)(6x - 5x + 1) = 0 ⇒ 
6x 2 − 5x + 1 = 0

 5 +1 6 1
x 2 = = ⇒ x2 =
5 ± 5 − 4·6·1 5 ± 1 
2
12 12 2
6x 2 − 5x + 1 = 0 ⇒ x = = =
6·2 12  5 − 1 4 1
 x 3 = 12 = 12 ⇒ x 3 = 3


4
7) x – 1 = 0

x 4 − 1 = 0 ⇒ x 4 = 1 ⇒ x = 4 1 = ± ⇒ x = ±1

3 2
8) 8x – 14x + 7x – 1 = 0
3 2 2
8 -14 7 -1 8x – 14x + 7x – 1 = (x + 2)(8x – 6x + 1)
1 8 -6 1
2
 x − 1 = 0 ⇒ x1 = 1
8 -6 1 0 (x – 1)(8x – 6x + 1) = 0 ⇒ 
8x − 6x + 1 = 0
2

 6+2 8 1
x 2 = = ⇒ x2 =
6 ± 6 − 4·8·1 6 ± 2  2
16 16 2
8x − 6x + 1 = 0 ⇒ x =
2
= =
8·2 16  6−2 4 1
 x 3 = 16 = 16 ⇒ x 3 = 4


4 3
9) 2x – 5x + 5x – 2 = 0
4 3 2
2 -5 0 5 -2 2x – 5x + 5x – 2 = (x + 1)( x – 1)(2x – 5x + 2)
1 2 -3 -3 2
 x − 1 = 0 ⇒ x1 = 1
2 -3 -3 2 0 
2 
(x + 1)( x – 1)(2x – 5x + 2) = 0 ⇒  x + 1 = 0 ⇒ x 2 = −1
-1 -2 5 -2  2
2 -5 2 0 2x − 5x + 2 = 0

 5+3 8
 x3 = = ⇒ x3 = 2
5 ± 5 − 4·2·2 5 ± 3 
2
4 4
2x 2 − 5x + 2 = 0 ⇒ x = = =
2·2 4 x = 5 − 3 2 1
= ⇒ x4 =
 4 4 4 2

4 2
10) x + 2x + 3 = 0

−2 ± 22 − 4·3
y2 + 2y + 3 = 0 ⇒ y = ⇒ No hay solución real
2

Luisa Muñoz 2