How do you solve #sqrt(x-2) + sqrt(x+3) = 5#?

1 Answer
Gió
Sep 15, 2015

I found #x=6#

Explanation:

We can try squaring both sides to get:
#(x-2)+2sqrt(x-2)sqrt(x+3)+(x+3)=25#
#x-2+2sqrt(x-2)sqrt(x+3)+x+3=25#
#2x+1+2sqrt(x-2)sqrt(x+3)=25#
rearranging:
#2sqrt(x-2)sqrt(x+3)=24-2x#
#sqrt(x-2)sqrt(x+3)=12-x#
square again:
#(x-2)(x+3)=144-24x+x^2#
#cancel(x^2)+3x-2x-6=144-24x+cancel(x^2)#
and:
#25x=150#
#x=6#

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