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

1 Answer
Shwetank Mauria
Feb 4, 2017

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#

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
4977 views around the world
You can reuse this answer
Creative Commons License