<s>[INST] 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!
 [/INST] Why did things get compilicated?
If you replace $x$ with $y^2$ and $\sqrt x$ with $y$ you should get
$y^2 - 8y +7 =0$ and that should not be complicated.
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Factor or use the quadratic formula so 
$(y - 7)(y-1) = 0$ so $y = 7$ or $y=1$.
Or $y = \frac {-(-8) \pm \sqrt {64-4*7*1}}2 = \frac {8\pm {36}}2 = \frac {8\pm 6}2=4\pm 3=1,7$; so $y=7$ or $y =1$.
So $x = y^2$ so $x = 7^2 = 49$ or $x = 1$.  Thus are the two solutions.
Check: 
If $x = 49$ then $x - 8\sqrt x +7 = 49 - 8*7 + 7 = 49- 56 + 7 = 0$.
And if $x = 1$ then $x - 8\sqrt x + 7 = 1-8*1 + 7 = 0$.
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