Number Subtracted to its Root

Let $f(x)=x-\sqrt{x}$ be defined for all $x\geq 0$ . Then $f'(x) = 1-\frac{1}{2\sqrt{x}}$ . Solving $f'(x)=0$ tells us that $x=\frac{1}{4}$ is the only critical point. Furthermore, since $f''\left(\frac{1}{4}\right) = \frac{1}{4\left(\frac{1}{4}\right)^{\frac{3}{2}}}=2>0$ we know that $x=\frac{1}{4}$ is a local minimum. Our function is differentiable for all $x>0$ , so the only other possible absolute minimum is $x=0$ . But $f(0)=0$ and $f\left(\frac{1}{4}\right)=-\frac{1}{4}$ . So $x=\boxed{\frac{1}{4}}$ is the value of $x$ that minimizes the given expression.

Since $x$ has to be positive, let $x=y^2$

The expression becomes $y^2-y$

Take the derivative and get $2y-1=0,y=0.5$

Hence $x=y^2=0.25$