$$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$
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?
Best Answer
You have to pay attention to your domain. In the first equation you get a positive and a negative value for $y$, while $\sqrt{x}$, which is your substitution, can only be positive.