How do you find the zeroes for #f(x)=2x^4-6x^2+1#?

1 Answer
Daniel L.
Jul 8, 2015

#x=+-sqrt((6-2sqrt(7))/4)# or #x=+-sqrt((6+2sqrt(7))/4)#

Explanation:

The first step is to substitute #t=x^2# which changes the degree of the equation to 2.
Now you get: #2t^2-6t+1=0# and you solve it as any other quadratic equation.

#Delta=6^2-4*2*1=36-8=28#
#sqrt(Delta)=2sqrt(7)#

#t_1=(6-2sqrt(7))/4#
#t_2=(6+2sqrt(7))/4#

Before we go further we must remember, that #t=x^2#, so #t# cannot be negative. Both values fulfill this condition, so we can ca;lculate #x# as:

#x_1=sqrt(t_1)#
#x_2=-sqrt(t_1)#
#x_3=sqrt(t_2)#
#x_4=-sqrt(t_2)#

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