How do you solve #\sqrt { 7- n } + \sqrt { n + 11} = 6#?

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Answer 1 · 2016-11-20T12:49:42

#n=-2#

The feasible values for #n# are

#7-n ge 0# and #n+11 ge 0# or #-11 le n le 7#

Making #7-n=y^2# we have

#y+sqrt(7-y^2+11)=6# or

#sqrt(7-y^2+11)=6-y# squaring both sides

#18-y^2=36-12y+y^2# or

#2(y^2-6y+9)=0# Solving for #y#

#y=(6pmsqrt(36-36))/2=3#

so #7-n=9# then #n = -2#