What are the solutions to the equation x^4-x^2=12 ?

The solutions to the equation x^4-x^2=12 are x=2,x=-2,x=sqrt(3)i,x=-sqrt(3)i

Find the zeros of x^4-x^2=12

The zeros of x^4-x^2=12 are x=2,x=-2,x=sqrt(3)i,x=-sqrt(3)i

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x^{4} - x^{2} = 12

x = 2, x = - 2, x = 3 i, x = - 3 i

Solution steps

x^{4} - x^{2} = 12

Move

12

to the left side

x^{4} - x^{2} - 12 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

u^{2} - u - 12 = 0

Solve

u^{2} - u - 12 = 0 : u = 4, u = - 3

u = 4, u = - 3

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = 4 : x = 2, x = - 2

Solve

x^{2} = - 3 : x = 3 i, x = - 3 i

The solutions are

x = 2, x = - 2, x = 3 i, x = - 3 i

3 q (q + 6) 2 (q + 6) = 0

3 x^{5} - 8 x^{4} + 2 x^{3} + 5 x^{2} - 2 x + 1 = 0

2 x^{3} - 6 x + 4 = 0

x^{4} + 9 x^{3} + 23 x^{2} + 3 x - 36 = 0

6 x^{5} + 11 x^{4} - 39 x^{3} - 41 x^{2} + 45 x + 18 = 0

What are the solutions to the equation x^4-x^2=12 ?
The solutions to the equation x^4-x^2=12 are x=2,x=-2,x=sqrt(3)i,x=-sqrt(3)i
Find the zeros of x^4-x^2=12
The zeros of x^4-x^2=12 are x=2,x=-2,x=sqrt(3)i,x=-sqrt(3)i