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How do you solve #sqrta-5=0#?

2 Answers

May 26, 2015

If #sqrt(a) - 5 =0#

then
#sqrt(a) = 5#

and
#a = 25#

May 27, 2015

#sqrt(a)-5=0#

#sqrt(a)=5#

#a^(1/2)=5#

#(a^(1/2))^2=5^2#

#a=5^2#

#a=25#

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