x+x+1=2x+1\sqrt{x} + \sqrt{x+1} = \sqrt{2x+1}x+x+1=2x+1
Squaring both sides, we have
(x+x+1)2=(2x+1)2\left(\sqrt{x} + \sqrt{x+1}\right)^{2} = \left(\sqrt{2x+1}\right)^{2}(x+x+1)2=(2x+1)2
x+2x(x+1)+x+1=2x+1x + 2\sqrt{x(x+1)} + x + 1 = 2x + 1x+2x(x+1)+x+1=2x+1
2x+1+2x(x+1)−(2x+1)=02x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 02x+1+2x(x+1)−(2x+1)=0
(2x(x+1))2=02⇒4(x(x+1))=0\left(2\sqrt{x(x+1)}\right)^{2} = 0^{2} \Rightarrow 4(x(x+1)) = 0(2x(x+1))2=02⇒4(x(x+1))=0
∴x(x+1)=0\therefore x(x+1) = 0∴x(x+1)=0
x= 0 or -1 x = \; \text{0 or -1}\;x=0 or -1