<s>[INST] Why is my solution incorrect for solving these quadratic equations? $$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable, 
$y = \sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get: 
$\displaystyle \frac{2}{y^{2}} - \frac{5}{y} = 1$
Multiplying both sides by $y^{2}$ I get: $2 - 5y = y^{2}$
So I have $y^{2} +5y-2=0$
Solving for y using completing the square, I get: 
$\displaystyle y = -\frac{5}{2} \pm \frac{\sqrt{33}}{2}$
So I should square this answer to get $x$ since $y^2 = x$
Then my answers are $\displaystyle y = \frac{58}{4} \pm \frac{10\sqrt{33}}{4}$
But this isn't the correct solution. 
Also for $\#10$ I do the same thing: 
Let $y = \sqrt n$ then $y^2 = n$
So I have $\displaystyle \frac{3}{y^2} - \frac{7}{y} -6 = 0$
Multiplying everything by $y^{2}$ gives me $3 -7y - 6y^2 = 0$
So I have $6y^{2} +7y - 3 = 0$
Solving for $y$ using the payback method I get: $\displaystyle y = -\frac{3}{2}, \frac{1}{3}$
Then $n = \frac{9}{4}, \frac{1}{9}$
But plugging these back in, my solution doesn't work.
I've been doing lots of quadratics using dummy variables and sometimes they work and sometimes they don't. 
Here are a list of my problems just so you have some reference: 
$$1)\ (x-7)^2 -13(x-7) +36=0 \qquad \qquad 4)\ 3(w/6)^2 -8(w/6) +4=0 \\
2)\ (1-3x)^2 -13(1-3x) +36=0 \qquad \qquad 5)\ 3(w^2-2)^2 -8(w^2-2) +4=0 \\
3)\ x^4 -13x^2 +36=0 \qquad \qquad 6)\ \frac{3}{p^2} -\frac{8}{p} +4=0$$
What am I doing wrong and how can I do these sorts of problems using dummy variables?
 [/INST] The other answers did not point out the logical gaps in your reasoning. It is unstated in your question but presumably you want to find non-negative real $x$ satisfying the given equation, so for any such $x$ you can let $y = \sqrt{x}$, and then you know that $y^2 = x$. Thus if you substitute $x$ by $y^2$ then solutions of the original equation are solutions of the new equation. But the new equation might have more solutions than the original!
Look at a simpler example first: If you want to solve $\sqrt{x} = x-2$ for real $x \ge 0$, squaring yields $x = (x-2)^2$ with solutions $x = 1$ or $x = 4$. But $1 = (x-2)^2$ does not imply that $1 = x-2$, so you should see why you cannot conclude that the original equation had the same solutions. It is also incorrect to just exclude the solutions based on 'domain', as this example shows clearly.
Symbolically, ( $x \in \mathbb{R}_{\ge0}$ and $\frac2x - \frac5{\sqrt{x}} = 1$ and $y = \sqrt{x}$ ) imply ( $x \in \mathbb{R}_{\ge0}$ and $\frac2x - \frac5{\sqrt{x}} = 1$ and $y^2 = x$ ) and hence imply ( $y \in \mathbb{R}$ and $\frac2{y^2} - \frac5y = 1$ and $y^2 = x$ ), but the reverse implications do not necessarily hold. The implications nevertheless allow you to conclude that every $x,y$ satisfying ( $x \in \mathbb{R}_{\ge0}$ and $\frac2x - \frac5{\sqrt{x}} = 1$ and $y = \sqrt{x}$ ) must also satisfy ( $y \in \mathbb{R}$ and $\frac2{y^2} - \frac5y = 1$ and $y^2 = x$ ), so you can limit your search to just the solutions of the latter, which are easier to find.
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