How do you find the zeroes for f(x)=2x^4-6x^2+1f(x)=2x4−6x2+1?

1 Answer
Daniel L.
Jul 8, 2015

x=+-sqrt((6-2sqrt(7))/4)x=±√6−2√74 or x=+-sqrt((6+2sqrt(7))/4)x=±√6+2√74

Explanation:

The first step is to substitute t=x^2t=x2 which changes the degree of the equation to 2.
Now you get: 2t^2-6t+1=02t2−6t+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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