[tex] f(x) = x^4-8x^2+5 [/tex]
We let x square =y
so x to the power 4 = y square
and we get
[tex] f(y)= y^2-8y+5 [/tex]
To find the zroes, we need to set f(y) to 0 and solve for y .
[tex] y^2 -8y+5=0 [/tex]
Now we use completing the square method
[tex] y^2 -8y +16=-5+16 [/tex]
[tex] (y-4)^2 = 11 [/tex]
[tex] y-4 = - \sqrt{11} , y-4 = \sqrt{11} [/tex]
[tex] y= 4 - \sqrt{11} , 4+ \sqrt{11} [/tex]
Now we do the back substitution.
[tex] x^2 = 4- \sqrt{11} \\ x = - \sqrt{4 - \sqrt{11}} , \sqrt{4-\sqrt{11}} [/tex] and
[tex] x^2 = 4+ \sqrt{11} \\ x = - \sqrt{4 + \sqrt{11}} , \sqrt{4+\sqrt{11}} [/tex]
