Question

How do you find the zeros of `y=x^4-7x^2+12`?

Answer 1

Zeros of `y=x^4-7x^2+12` are `-2,-sqrt3,sqrt3` and `2`

Explanation:

Zeroa of `y=x^4-7x^2+12` will be given by the solution to the equation `x^4-7x^2+12=0`. To solve this assume `u=x^2`, then the equation becomes

`u^2-7u+12=0`, which can be factorized as

`(u-4)(u-3)=0`

i.e. either `u=4` i.e. `x^2=4` or `x^2-4=0` or `(x+2)(x-2)=0` and `x=-2` or `2`.

or `u=3` i.e. `x^2=3` or `x^2-3=0` or `(x+sqrt3)(x-sqrt3)=0` and `x=-sqrt3` or `sqrt3`.

Hence, zeros of `y=x^4-7x^2+12` are `-2,-sqrt3,sqrt3` and `2`