What are the solutions to the equation (x^2+3x)^2-(2x^2+6x)-8=0 ?

The solutions to the equation (x^2+3x)^2-(2x^2+6x)-8=0 are x=-1,x=1,x=-2,x=-4

Find the zeros of (x^2+3x)^2-(2x^2+6x)-8=0

The zeros of (x^2+3x)^2-(2x^2+6x)-8=0 are x=-1,x=1,x=-2,x=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^{2} + 3 x)^{2} - (2 x^{2} + 6 x) - 8 = 0

x = - 1, x = 1, x = - 2, x = - 4

Solution steps

(x^{2} + 3 x)^{2} - (2 x^{2} + 6 x) - 8 = 0

Expand

(x^{2} + 3 x)^{2} - (2 x^{2} + 6 x) - 8 : x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8

x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8 = 0

Factor

x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8 : (x + 1) (x - 1) (x + 2) (x + 4)

(x + 1) (x - 1) (x + 2) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 1 = 0 or x + 2 = 0 or x + 4 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

Solve

x + 4 = 0 : x = - 4

The solutions are

x = - 1, x = 1, x = - 2, x = - 4

x^{12} = 7

x^{12} = 8

5 x^{3} - 2 x^{2} - 5 x + 2 = 0

(2 + 3) \cdot x^{4} = 20

15 x^{4} - 20 x^{3} = 0

What are the solutions to the equation (x^2+3x)^2-(2x^2+6x)-8=0 ?
The solutions to the equation (x^2+3x)^2-(2x^2+6x)-8=0 are x=-1,x=1,x=-2,x=-4
Find the zeros of (x^2+3x)^2-(2x^2+6x)-8=0
The zeros of (x^2+3x)^2-(2x^2+6x)-8=0 are x=-1,x=1,x=-2,x=-4