How do you solve #a^4-1=0#?

2 Answers
Veddesh #phi#
Jan 6, 2016

#a=1,-1#

Explanation:

#a^4-1=0#

#a^4=1#

The only solutions of a could be 1 and -1 because,

#1*1*1*1=1#

#-1*-1*-1*-1=1#

Lucio Falabella
Jan 6, 2016

#a=+-1#

Explanation:

alternatively

#(a^4-1)=(a^2+1)times(a^2-1)=(a^2+1)times(a+1)times(a-1)=0#
Now a moltiplation is zero when the factors are zero
#(a^2+1)=0# #cancel(EE) x in RR# (no Real solutions)
#(a+1)=0#; #a=-1#
#(a-1)=0#; #a=+1#

Related questions
See all questions in Comparing Methods for Solving Quadratics
Impact of this question
2205 views around the world
You can reuse this answer
Creative Commons License