Question 427364


{{{sqrt(x+1) +1=sqrt(2x)}}}.......to power of 2


{{{(sqrt(x+1))^2 +2*sqrt(x+1)+1=(sqrt(2x))^2}}}



{{{x+1 +2*sqrt(x+1)+1=2x}}}

{{{2*sqrt(x+1)=2x-x-1-1}}}

{{{2*sqrt(x+1)=x-2}}}...to power of 2

{{{(2*sqrt(x+1))^2=(x-2)^2}}}

{{{4(x+1)=x^2-4x+4}}}

{{{4x+4=x^2-4x+4}}}

{{{0=x^2-4x-4x-4+4}}}

{{{0=x^2-8x}}}

{{{0=x(x-8)}}}


so, x=0...if that, we will have no solution because {{{sqrt(2x)}}} 


or  {{{(x-8)=0}}}..=>..{{{x=8}}}




check:


if {{{x=8}}}

{{{sqrt(8+1) +1=sqrt(2*8)}}}

{{{sqrt(9) +1=sqrt(16)}}}


{{{3 +1=4}}}

{{{4=4}}}