What is the answer to 3=sqrt(3x^3+2x^2+3x+1) ?

The answer to 3=sqrt(3x^3+2x^2+3x+1) is x=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3=sqrt(3x^3+2x^2+3x+1)

3 = 3 x^{3} + 2 x^{2} + 3 x + 1

x = 1

Solution steps

3 = 3 x^{3} + 2 x^{2} + 3 x + 1

Square both sides:

9 = 3 x^{3} + 2 x^{2} + 3 x + 1

9 = 3 x^{3} + 2 x^{2} + 3 x + 1

Solve

9 = 3 x^{3} + 2 x^{2} + 3 x + 1 : x = 1

x = 1

Verify Solutions:

x = 1

True

The solution is

x = 1

2 x + 1 + x - 1 = 1

2 = x^{\frac{5}{3}}

x^{2} + 3 x - x^{2} + 3 x - 1 - 7 = 0

x + 10 + x + 19 = - 1

(x - 5)^{\frac{3}{2}} = 64

What is the answer to 3=sqrt(3x^3+2x^2+3x+1) ?
The answer to 3=sqrt(3x^3+2x^2+3x+1) is x=1
Solve for x: 3=sqrt(3x^3+2x^2+3x+1)
The solution is x=1