$$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$

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?

  • $\begingroup$ You should explain how you check whether your results are correct. $\endgroup$ – Carsten S Nov 25 '18 at 15:00
  • $\begingroup$ Why not $y = \frac{1}{\sqrt{x}}$? $\endgroup$ – Eric Towers Nov 25 '18 at 17:05

Based on comments some of your issues seemed to be in checking answers and others likely an issue in arithmetic (even the greatest sometimes think $1+1=3$ every now and then, but I can’t speak for them). Here’s a tip I have for checking answers to quadratics.

Generally you have something of the form:

$$Ax^2 + Bx + C = 0$$

You only need to know it is zero. So what I did was this. I didn’t multiply everything. I plugged enough in to verify. So say D = x. I might do the following:

$$DAx + Bx + \frac CD x = 0$$

Now you just take the sum $DA + B + \frac CD$ which can be done mentally with most high school practice problems.

In the case of $(x-7)^2 - 13(x-7) + 36 = 0$ this makes it much easier to verify $11$ and $16$ as answers.

As for general quadratics your style and technique is good. I don’t see any flaws in your thinking aside from being careful about the back substitutions. Definitely watch for dropping negatives and adding them in. Doing your own new problem might introduce needs to carefully think about the Complex Plane if you aren’t careful. However picking the right substitutions is an art and one I cannot really describe with a process or answer. Most people at your level do not substitutions and so your definitely thinking outside the box. This is a good thing. I did not even know if substitutions until Calculus and it made it very hard to grasp that concept (of just creating a new variable on the fly) so kudos to you.

  • $\begingroup$ Your equation "$DAx+Bx+\frac{C}{D}x = 0$" makes no sense, so I don't know how you got the (correct) "$DA+B+\frac{C}{D} = 0$" out of that. $\endgroup$ – user21820 Nov 25 '18 at 4:47
  • $\begingroup$ @user21820 the wording is a little confusing, but if $x=D$ then it is indeed the case that $Ax^2+Bx+C=DAx+Bx+\frac{C}{D}x$. $\endgroup$ – Carmeister Nov 25 '18 at 5:33
  • $\begingroup$ @Carmeister: You are right! I guess my brain just balked at the "$x$", since we cannot divide by $0$, and didn't think further. $\endgroup$ – user21820 Nov 25 '18 at 5:40
  • $\begingroup$ @user21820 indeed but the case that $x = 0$ is so easy to plug in to verify that it need not be mentioned. $\endgroup$ – The Great Duck Nov 25 '18 at 6:11
  • $\begingroup$ @TheGreatDuck: Agreed. I didn't downvote, but perhaps you should clarify your post by splitting cases as you stated in your comment and completely substituting $x = D$ before dividing by $D$. Note that you're not actually addressing the question, but for some reason the asker accepted your answer, so you might as well improve it. $\endgroup$ – user21820 Nov 25 '18 at 6:28

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.

  • $\begingroup$ even if i choose the positive value for x and try to plug it back in, it isn't the correct answer $\endgroup$ – user130306 Nov 25 '18 at 2:56
  • 1
    $\begingroup$ Not sure how you checked your answers, but one of them is surely correct, namely the one with the minus in the first equation. $\endgroup$ – Makina Nov 25 '18 at 2:57
  • $\begingroup$ actually I just redid my question 1, (which is on the bottom) and there, there is no restriction on the domain since x can be anything, and i still get teh wrong answer. I get $x =11, 16$ and that doesn't work. I'm referencing $(x-7)^{2}-13(x-7)+36=0$ btw $\endgroup$ – user130306 Nov 25 '18 at 2:58
  • $\begingroup$ @user130306 x = 11 is definitely a solution by mental math. Checking 16 now. $\endgroup$ – The Great Duck Nov 25 '18 at 2:59
  • $\begingroup$ so you're saying $x = \frac{58}{4} - \frac{10\sqrt 33}{4}$ works? $\endgroup$ – user130306 Nov 25 '18 at 2:59

You are doing your algebra correctly, the only problem is when you have a positive and a negative value for a square root you have to ignore the negative one.

I checked the answer $$ x= \frac {58-10 \sqrt {33}}{4}$$ for your first problem and it does work nicely.


In your case, substituting $x = y^2$ might introduce extraneous roots. So you need to check that your computed answers are really answers.

Without any substitution, you could write $\dfrac2x -\dfrac5{\sqrt{x}}=1$ as

$$2 \left(\dfrac{1}{\sqrt x}\right)^2 - 5\left(\dfrac{1}{\sqrt x}\right)-1 = 0$$

So $$\dfrac{1}{\sqrt x} = \dfrac{5 \pm \sqrt{33}}{4}$$

We can ignore $\dfrac{1}{\sqrt x} = \dfrac{5 - \sqrt{33}}{4}$ since the number on the right side is negative.

So we get $$\sqrt x = \dfrac{4}{5+\sqrt{33}}= \dfrac{\sqrt{33}-5}{2}$$

and $$x = \dfrac{29-5\sqrt{33}}{2}$$

Problem $(5)$ for example, can be written as

\begin{align} 3\color{red}{(w^2-2)}^2 -8\color{red}{(w^2-2)} + 4 &= 0 \\ (3\color{red}{(w^2-2)} - 2)(\color{red}{(w^2-2)} - 2) &= 0 \\ w^2-2= \dfrac 23 &\text{ or } w^2-2 = 2 \\ w^2 = \dfrac 83 &\text{ or } w^2 = 4 \\ w &\in \left\{\dfrac 23 \sqrt 6, -\dfrac 23 \sqrt 6, 2, -2 \right\} \end{align}

You can solve the others similarly.


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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