Hello!

Im struggling with finding the solutions for: \(x^4+2x^2-3=0\)

Thankfull for help :D

kilander Nov 21, 2017

#1**+2 **

Solve for x:

x^4 + 2 x^2 - 3 = 0

Substitute y = x^2:

y^2 + 2 y - 3 = 0

The left hand side factors into a product with two terms:

(y - 1) (y + 3) = 0

Split into two equations:

y - 1 = 0 or y + 3 = 0

Add 1 to both sides:

y = 1 or y + 3 = 0

Substitute back for y = x^2:

x^2 = 1 or y + 3 = 0

Take the square root of both sides:

x = 1 or x = -1 or y + 3 = 0

Subtract 3 from both sides:

x = 1 or x = -1 or y = -3

Substitute back for y = x^2:

x = 1 or x = -1 or x^2 = -3

Take the square root of both sides:

**x = 1 or x = -1 or x = i sqrt(3) or x = -i sqrt(3)**

Guest Nov 21, 2017

#1**+2 **

Best Answer

Solve for x:

x^4 + 2 x^2 - 3 = 0

Substitute y = x^2:

y^2 + 2 y - 3 = 0

The left hand side factors into a product with two terms:

(y - 1) (y + 3) = 0

Split into two equations:

y - 1 = 0 or y + 3 = 0

Add 1 to both sides:

y = 1 or y + 3 = 0

Substitute back for y = x^2:

x^2 = 1 or y + 3 = 0

Take the square root of both sides:

x = 1 or x = -1 or y + 3 = 0

Subtract 3 from both sides:

x = 1 or x = -1 or y = -3

Substitute back for y = x^2:

x = 1 or x = -1 or x^2 = -3

Take the square root of both sides:

**x = 1 or x = -1 or x = i sqrt(3) or x = -i sqrt(3)**

Guest Nov 21, 2017