Burgers equation with initial data $u(x,0) = x^2$

Question

I have the PDE

$$u_t + u u_x = 0, t>0$$

$$u(x,0) = f_0(x) = x^2$$

Reading this answer we arrive at the solution

$$u(x, t) = f_0(x-ut) = (x-ut)^2$$

$$u = x^2 - 2xut + u^2 t^2 =0$$

$$u^2 t^2 -u(1+2xt) + x^2=0$$

Which gives, using the quadratic formula,

$$u(x, t) = \frac{(2xt+1) \pm \sqrt{(2xt +1)^2 - (2xt)^2}}{2t^2}$$

However this does not agree with $u = x^2$ as $t$ approaches $0$.

Is there some mistake I have made?

Answer 1

$$ \lim_{t\to0}\frac{(2\,x\,t+1)-\sqrt{(2\,x\,t +1)^2-(2\,x\,t)^2}}{2\,t^2}=\lim_{t\to0}\frac{2\,x^2}{(2\,x\,t+1)+\sqrt{(2\,x\,t +1)^2-(2\,x\,t)^2}}=x^2. $$