How do you solve #sqrt(x+2) + 1 =sqrt(3-x)#?

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Answer 1 · 2017-02-28T17:06:12

#x=2-x-2sqrt(3-x)#

#sqrt(x+2)+1=sqrt(3-x)#

Let #sqrt(x+2)=a#

Let #sqrt(3-x)=b#

#:.sqrta+1=sqrtb#

#:.sqrta=sqrtb-1#

#a=(sqrtb-1)^2#

#:.a=(sqrtb-1)(sqrtb-1)#

#:.a=(sqrtb)^2-2sqrt2+1#

#:.a=b-2 xx (2)^(1/2)+1#

#:.a=b-(4)^(1/2)+1#

#:.a=b-(2^2)^(1/2)+1#

#:.a=b-2+1#

#:.a=b-1#

substitute #a=sqrt(x+2)# and #b=sqrt(3-x)#

#:.sqrt(x+2)=sqrt(3-x)-1#

#:.x+2=(sqrt(3-x)-1)^2#

#:.x+2=(sqrt(3-x))^2-2(sqrt(3-x))+1#

#:.x+2=3-x-2sqrt(3-x)+1#

#:.x=4-2-x-2sqrt(3-x)#

#:.x=2-x-2sqrt(3-x)#

Answer 2 · 2017-02-28T17:29:34

#x = {-1, 2}#

The feasible solutions are those verifying #x+2 ge 0# and #3-x ge 0# or #-2 le x le 3#

Now #sqrt(x+2) = sqrt(3-x) - 1# squaring both terms

#x+2=3-x-2sqrt(3-x)+1# or

#2x-2 = 2 sqrt(3-x)->x-1=sqrt(3-x)#

squaring again

#x^2-2x+1=3-x->x^2-x-2=0# with solution

#x = (1 pm 3)/2-> (x = -1, x = 2)# Testing now for feasibility

#sqrt(-1+2)=sqrt(3-(-1))-1# and
#sqrt(-2+2)=sqrt(3-2)-1#

so both solutions are feasible.