How do you find the zeros, if any, of #y= -x^4 -3x^2+17 #using the quadratic formula?

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Answer 1 · 2016-02-18T22:14:23

#+-sqrt((-3+sqrt(77))/(2))#

if #a=x^2# you will obtain:

#-a^2-3a+17#

Apply the second degree formula:

#(3+-sqrt(3^2-4xx(-1)xx17))/(2xx(-1))#

#(3+-sqrt(9+68))/(-2)#

#(3+-sqrt(77))/(-2)#

#(-3+-sqrt(77))/(2)#

We know that #x=sqrt(a)#

So the real solutions will be:

#+-sqrt((-3+sqrt(77))/(2))#