Question

How do you solve `4x^4-21x^2+27=0`?

Answer 1

`x = +-3/2 and x = +-sqrt3`

Explanation:

`4x^4 -21x^2 +27 =0`

This is a disguised quadratic. Note that `(x^2)^2 = x^4`

You need to find factors first.

Find factors of 4 and 27 which cross-multiplied and ADDED give 21.

[As 21 is an odd number, it has to come from even + odd,
so the factors of 4 CANNOT be `2xx2`, because 2 will always give an even product, and even + even = even]

`""4" "27`
`darrcolor(white)(....)darr`

`4color(white)(........)9" "rarr 1xx9 = 9`
`1color(white)(........)3" "rarr 4xx3 = ul12`
`color(white)(......................................)21`

The signs in the bracket will both be negative (because of `-21`)

`4x^4 -21x^2 +27 = 0`

`rarrcolor(blue)((4x^2-9))(x^2-3)=0" "larr` difference of squares

`rarrcolor(blue)((2x+3)(2x-3))(x^2-3)=0" "larr` there are 3 factors

Set each factor equal to 0 and solve.

`2x+3 = 0 " "rarr x = -3/2`
`2x-3 = 0 " "rarr x = 3/2`
`x^2-3 = 0 " "rarr x = +-sqrt3`

We have 4 solutions as we expect for a fourth order equation.