Question

Solve for real x: ${\left(\sqrt{x}\right)}^2+2\sqrt{x}+1=0$

Answer 1

Given ${\left(\sqrt{x}\right)}^2$$+2$$\sqrt{x}$$+1=0$

${\left(\sqrt{x}\right)}^2$$+$$\sqrt{x}$$+$$\sqrt{x}$$+1=0$

$\sqrt{x}$($\sqrt{x}$$+1$)$+1$($\sqrt{x}$$+1$)$=0$

($\sqrt{x}$$+1$)($\sqrt{x}$$+1$)$=0$

$\sqrt{x}$ $=$ $-1$

but the square root cannot be negative.

So , No Solution.