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Hello!

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

 

Thankfull for help :D

kilander  Nov 21, 2017

Best Answer 

 #1
avatar
+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
avatar
+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

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