solve the polynomial equation?

Question

x^4 - 4x^2 - 5 = 0 ..?

Answer 1

To do this one, make a substutituion y = x^2 and solve
y^2 - 4y - 5 = 0 Then when you find y, x = +/- sqrt y.
This will give you up to four answers. here we go...
Factor
(y-5)(y+1) = 0
So by the zero factor theorem
y-5 = 0 or y + 1 = 0
y = 5 or y = -1
So x = +/- sqrt 5 or +/- 1i (where i is the number whose square is -1)
If you only want real number solutions, discard the +/- i and say x = +/- sqrt 5

Answer 2

First: factor the expression. Multiply the 1st and 3rd equation to get - 5. Find two numbers that give you - 5 when multiplied and
- 4(2nd-middle coefficient) when added/subtracted. The numbers are: - 5, 1

Sec: rewrite the expression with the new middle coefficients...

x^4 - 5x^2 + x^2 - 5 = 0

*When you have 4 terms - group "like" terms...

(x^4 + x^2) - (5x^2 - 5) = 0

*Factor both sets of parenthesis...

x^2(x^2 - 1) - 5(x^2 - 1) = 0
(x^2 - 1)(x^2 - 5) = 0

*Factor "x^2 - 1" >>>

(x + 1)(x - 1)(x^2 - 5) = 0

Third: solve the x-variables > set the parenthesis to equal "0" >

1. x + 1 = 0
x + 1 - 1 = 0 - 1
x = -1

2. x - 1 = 0
x - 1 + 1= 0 + 1
x = 1

3. x^2 - 5 = 0

*Add 5 to both sides...

x^2 - 5 + 5 = 0 + 5
x^2 = 5

*Eliminate the exponent > find the square root of both sides...

V`x^2 = +/- V`5
x = V`5, -V`5

x = - 1, 1, V`5, -V`5

Answer 3

Let y=x^2

y^2 - 4y -5 =0
(y-5)(y+1)=0
y=5,-1
x^2 = y
x= +/- sqrt(5), +/- i

where i is the sqrt of -1, the base of the imaginary number system.

Answer 4

This is a simple trinomial:

(x^2 - 5)(x^2 + 1) = 0
x^2 = 5 or x^2 = -1
x = sqrt5 or x = -sqrt5 the next two only exist in complex numbers: x = i or x = -i

Answer 5

i got -5/4x^2

Answer 6

x={-2.236068, 2.236068}