How do you solve #3+sqrt[x+7]=sqrt[x+4]# and find any extraneous solutions?

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Answer 1 · 2016-06-16T17:30:36

the equation is impossible

you can calculate

#(3+sqrt(x+7))^2=(sqrt(x+4))^2#

#9+x+7+6sqrt(x+7)=x+4#

that's

#6sqrt(x+7)=cancel(x)+4-9cancel(-x)-7#

#6sqrt(x+7)=-12#

that's impossible because a square root must be positive

Answer 2 · 2016-06-16T17:55:28

No real roots of #x# exist in #R# (#x!inR#)

#x# is a complex number #x=4*i^4-7#

First to solve this equation we think how to take off the square root, by squaring both sides:

#(3+sqrt(x+7))^2=(sqrt(x+4))^2#

Using the binomial property for squaring of sum
#(a+b)^2=a^2+2ab+b^2#

Applying it on both sides of the equation we have:

#(3^2+2*3*sqrt(x+7)+(sqrt(x+7))^2)=x+4#

Knowing that #(sqrt(a))^2=a#
#9+6sqrt(x+7)+x+7=x+4#

Taking all the know a and unknowns to the second side leaving the square root on one side we have:

#6sqrt(x+7)=x+4-x-7-9#
#6sqrt(x+7)=-12#
#sqrt(x+7)=-12/6#
#sqrt(x+7)=-2#

Since square root equal to a negative real number that is
impossible in #R#, no roots exists so we have to check complex set.

#sqrt(x+7)=-2#

Knowing that i^2=-1 that means #-2=2*i^2#

#sqrt(x+7)=2i^2#
Squaring both sides we have:

#x+7=4*i^4#
Therefore , #x=4*i^4-7#

So #x # is a complex number.