#sqrt( x^2 + 5) = x + 1? #solve the equation

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Answer 1 · 2017-04-13T19:45:13

#x=2#

#sqrt(x^2+5)=x+1#

#x^2+5=(x+1)^2=x^2+2x+1#

#x^2+5=x^2+2x+1#

#2x-4=0#

#x=2#

Verify solution using original equation

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

#2+1=3#

#x=2# is a valid solution