What are the solutions of the equation x^4+4x^3+2x^2-4x-m=0 and what is the value of m if equation has real solutions?

1 Answer | Add Yours

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write the equation into an equivalent form:

x^2(x+2)^2 - 2x(x+2) - m = 0

We'll substitute the product x(x+2) by t:

t^2 - 2t - m = 0

We'll apply the quadratic formula:

t1 = [2 + sqrt(4 + 4m)]/2

t1 = [2 + 2sqrt(1+m)]/2

t1 = 1 + sqrt(1+m)

t2 = 1 - sqrt(1+m)

But t = x(x+2)

We'll remove the brackets:

x^2 + 2x - t1 = 0

x^2 + 2x - 1 - sqrt(1+m) = 0

x1 = -1+sqrt[2+sqrt(1+m)]

x2 = -1-sqrt[2+sqrt(1+m)]

x^2 + 2x - t2 = 0

x^2 + 2x - 1 + sqrt(1+m) = 0

x3 = -1+sqrt[2-sqrt(1+m)]

x4 = -1-sqrt[2-sqrt(1+m)]

The roots x1,x2,x3 and x4 are real if and only if the m satisfies the constraints:

1+m>=0 => m>=-1

2 - sqrt(1+m) >= 0

m belongs to [-1 ; +infinite) and (-infinite ; 3]

For the equation to have all real solutions, m = [-1 ; 3].

We’ve answered 318,916 questions. We can answer yours, too.

Ask a question