# solve x :  x^4 -4x^2 -32 = 0

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

x^4 - 4x^2 - 32 = 0

Let us assume that x^2 = y

==> y^2 - 4y -32 = 0

Now we can factorize:

==> (y-8)(y+4)

==> y1= 8 => 8= x1^2 ==> x1= +-sqrt8 = +-2sqrt2

==> y2= -4 ==: -4 = x^2 (impossible for real numbers )

For complex number:    ==> x = +-2i

Then the solution is:

x1= 2sqrt2

x2= -2sqrt2

x3= 2i

x4= -2i

kjcdb8er | Teacher | (Level 1) Associate Educator

Posted on

x^4 -4x^2 -32 = 0
(x^2-8) (x^2+4) = 0

(x+sqrt(8))(x-sqrt(8)(x+2i)(x-2i)

x = 2√2, -2√2, 2i, -2i

neela | High School Teacher | (Level 3) Valedictorian

Posted on

x^4-4x^2-32 = 0 To solve for x.

Solution:

This is a quadratic equation in x^2. So we solve it  by factoring as a quadratic.

t^2-4t-32 = 0, where t= x^2

t^2-8t+4t-32 = 0

t(t-8)+4(t-8) = 0

(t-8)(t+4) = 0

t-8 = 0 or t+4 = 0

t-8= 0 gives x^2 = 8 , x = +or- sqrt8 = 2sqrt2 or -2sqrt2

t+4 = 0 gives x = sqrt(-4) = +or-2i

So the solution set is { -2i,2i, -2srt2, 2srt2}