Friday, May 10, 2013

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

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}.

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