# Find all solutions of x^4 - 3x^2 + 2 = 0?

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We have to find all the solutions of x^4 - 3x^2 + 2 = 0.

Now let y = x^2

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

=> y^2 - 3y + 2 = 0

=> y^2 - 2y - y + 2 =0

=> y*(y - 2) - 1*(y - 2) =0

=> (y - 1)(y - 2) = 0

=> y = 1 or y = 2

As y = x^2

x^2 = 1

=> x = 1 or x = -1

x^2 = 2

=> x = sqrt 2 or -sqrt 2

**Therefore the solutions of the equation x^4 - 3x^2 + 2 = 0 are 1, -1, sqrt 2, -sqrt 2.**

The equation will have 4 solutions since it's order is 4.

We'll use factorization to solve it. For the beginning, we'll re-write the middle terms as a sum of 2 terms:

-3x^2 = -x^2 - 2x^2

We'll substitute the middle terms by the algebraic sum:

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

We'll group the first 2 terms and the last 2 terms:

(x^4 -x^2) - (2x^2 - 2) = 0

We'll factorize by x^2 the first group and by 2 the last group;

x^2(x^2 - 1) - 2(x^2 - 1) = 0

We'll factorize by (x^2 - 1):

(x^2 - 1)(x^2 - 2) = 0

We'll set each factor as zero:

x^2 - 1 = 0

We'll re-write the difference of squares, using the formula:

a^2 - b^2 = (a-b)(a+b)

We'll put a = x and b = 1

x^2 - 1 = (x-1)(x+1)

x - 1 = 0

x1 = 1

x + 1 = 0

x2 = -1

We'll do the same with the second factor x^2 - 2:

x^2 - 2 = (x - sqrt2)(x + sqrt2)

x - sqrt2 = 0

x3 = sqrt2

x4 = -sqrt2

** The 4 roots of the equation are: {-sqrt2 ; -1 ; 1 ; sqrt2}.**