Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!

If
$x - 8\sqrt x + 7 = 0, \tag 1$
then
$x + 7 = 8\sqrt x; \tag 2$
then
$x^2 + 14 x + 49 = (x + 7)^2 = (8\sqrt x)^2 = 64x; \tag 3$
thus
$x^2 - 50x + 49 = 0, \tag 4$
which factors as
$(x - 1)(x - 49) = 
x^2 - 50x + 49 = 0; \tag 5$
thus
$x = 1 \; \text{or} \; x = 49; \tag 6$
it is now a simple matter to check that $1$, $49$ obey (1).
