How do you solve using the quadratic formula #x^4 - 26x^2 +25#?

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Answer 1 · 2015-04-30T17:57:03

In this way:

let #t=x^2#, so:

#t^2-26t+25=0rArr#

#Delta=b^2-4ac=26^2-4*1*25=676-100=576=24^2#

#t_(1,2)=(-b+-sqrtDelta)/(2a)=(26+-24)/2#

#t_1=(26-24)/2=1#

#t_2=(26+24)/2=50/2=25#.

So:

#x^2=1rArrx=+-1#

#x^2=25rArrx=+-5#.

Since the coefficient #b# of the quadratic equation is even, we could use also the reducted formula.

#Delta/4=(b/2)^2-ac=169-25=144=12^2#

#t_(1,2)=(-b/2+-sqrt(Delta/2))/a=(13+-12)/1#

with, obviously, the same two solutions!