Find all solutions of x^4 - 3x^2 + 2 = 0?
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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}.