What is the answer to (x+2)^2-sqrt(x^2+4x-1)=7 ?

The answer to (x+2)^2-sqrt(x^2+4x-1)=7 is x=1,x=-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x+2)^2-sqrt(x^2+4x-1)=7

(x + 2)^{2} - x^{2} + 4 x - 1 = 7

x = 1, x = - 5

Solution steps

(x + 2)^{2} - x^{2} + 4 x - 1 = 7

Expand

(x + 2)^{2} - x^{2} + 4 x - 1 : x^{2} + 4 x + 4 - x^{2} + 4 x - 1

x^{2} + 4 x + 4 - x^{2} + 4 x - 1 = 7

Remove square roots

x^{2} + 4 x - 1 = x^{4} + 8 x^{3} + 10 x^{2} - 24 x + 9

Solve

x^{2} + 4 x - 1 = x^{4} + 8 x^{3} + 10 x^{2} - 24 x + 9 : x = 1, x = - 5, x = - 2 + 6, x = - 2 - 6

x = 1, x = - 5, x = - 2 + 6, x = - 2 - 6

Verify Solutions:

x = 1

True

, x = - 5

True

, x = - 2 + 6

False

, x = - 2 - 6

False

The solutions are

x = 1, x = - 5

x^{0.2} + 5 = 25

5 = 0.7 a^{0.3}

6 t + 4 = - 4

3 x + 2 - 2 x + 3 - 5 = 0

- 2 x - 3 + 1 = x

What is the answer to (x+2)^2-sqrt(x^2+4x-1)=7 ?
The answer to (x+2)^2-sqrt(x^2+4x-1)=7 is x=1,x=-5
Solve for x: (x+2)^2-sqrt(x^2+4x-1)=7
The solution is x=1,x=-5