Question

What are the real or imaginary solutions of the polynomial equation x^4-52x^2+576=0?

Answer 1

rewriting x^4 as A^2 and x^2 and A.

A^2-52A+576=0
(A-36)(A-16)=0
A=36,16

since A is x^2,

x^2=36 and x^2=16
x=6 and -6 and x=4 and -4

Answer 2

ANSWER

The solutions are,


`x = - 6 \: or \: x = 6 \: or \: x = - 4 \: or \: x = 4`


Step-by-step explanation

The given polynomial equation is,


`{x}^(4) - 52 {x}^(2) + 576 = 0`

We can rewrite the equation to obtain,




`( {x}^(2) )^(2) - 52 {x}^(2) + 576 = 0`

If we let


`u = {x}^(2)`

Then our equation becomes,



`{u}^(2) - 52u + 576 = 0`


This is a quadratic equation that can be solved by factoring.


We split the middle term to obtain,



`{u}^(2) - 16u - 36u+ 576 = 0`




This factors to give us,



`u(u - 16) - 36(u - 16) = 0`



`(u - 16)(u - 36) = 0`


`u = 16 \: or \: u = 36`

This implies that,


`{x}^(2) = 16 \:or \: {x}^(2) = 36`


`x = \pm4 \: or \: x = \pm6`

There real solutions are,


`x = - 6 \: or \: x = 6 \: or \: x = - 4 \: or \: x = 4`