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

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Answer 1 · 2016-07-14T01:27:58

Square both sides of the equation.

#(sqrt(x^2 + 5))^2 = (x + 3)^2#

#x^2 + 5 = x^2 + 6x + 9#

#x^2 - x^2 - 6x + 5 - 9 = 0#

#-6x = 4#

#x = -2/3#

Checking in the original equation, we find this solution works. Hence, the solution set is #{-2/3}#.

Hopefully this helps!