How do you simplify #sqrt((a +2)^2)#?
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"Which following pairs of atoms, have a lower electron affinity? a) Ca,K b) I,F c) Li, Ra. I seriously don't know anything about electron affinity all ik that it can buy another element"
#sqrt((a+2)^2)# means to take the square root of #(a+2)^2#, which is #a+2#.
#sqrt ((a+2)^2)=a+2##" "a in [-2,-oo)#
#" "#
#sqrt((a+2)^2) = -(a+2)# #" "a in (-oo,-2)#
#sqrt ((a+2)^2)=abs (a+2)#
#" "#
If#" "color (blue)(a+2>=0)rArra >= -2" "#
#" " #
then
#" "#
#sqrt ((a+2)^2)=color (blue)(a+2)# #" "a in [-2,-oo)#
#" "#
If#" "color (red)(a+2 < 0 ) rArr a <-2" "#
#" "#
then
#" "#
#sqrt((a+2)^2) = color (red)(-(a+2))# #" "a in (-oo,-2)#
#sqrt((a+2)^2) = abs(a+2)#
A square root of a number #x# is a number #y# such that #y^2 = x#.
Any non-zero number #x# has two square roots, which we write as #sqrt(x)# and #-sqrt(x)#. The principal square root is #sqrt(x)#.
If #x# is positive then #sqrt(x)# is the positive square root and #-sqrt(x)# the negative one.
If #x = t^2# for some number #t# then the square roots of #x# are #t# and #-t#.
Hence we find that the square roots of #(a+2)^2# are #(a+2)# and #-(a+2)#.
Which of #(a+2)# and #-(a+2)# is the principal, non-negative one? Whichever is positive, or if zero, then they are both the same.
We can automatically choose between the two using the absolute value and write:
#sqrt((a+2)^2) = abs(a+2)#
