I am trying to simplify this expression $$\frac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}= $$ However, I am not certain if I am doing it right or if I am simplifying it far enough. The end doesn't really seem right to me. If someone can confirm or correct me, it will be greatly appreciated. P.S I am very new to mathjax so I apologize if it is not as tidy as it should be. If someone can give me some advice on how to make it more appropriate, you have my appreciation.

\begin{align*} &= \frac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}\\ &= \left ( \frac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}} \right )\left ( \frac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}+\sqrt{x+1}} \right )\\ &= \frac{\left ( \sqrt{x} \right )^2 + 2\sqrt{x}\sqrt{x+1} + \left ( \sqrt{x+1} \right )^2}{\left ( \sqrt{x} \right )^2 - \left ( \sqrt{x+1} \right )^2} \\ &= \frac{x+2 \sqrt{x} \sqrt{x+1} + x + 1 }{ x-x-1}\\ &= \frac{ 2x + 2 \sqrt{x} \sqrt{x+1} + 1}{ -1}\\ &= -2x -2 \sqrt{x \left ( x+1 \right )} -1 \end{align*}

  • $\begingroup$ Looks good to me. $\endgroup$
    – user137731
    Oct 24, 2015 at 1:29
  • $\begingroup$ This is correct. And you are seeing a negative answer because the denominator is negative for the original expression $\endgroup$ Oct 24, 2015 at 1:32
  • $\begingroup$ One last thing: we don't "solve" expressions -- we evaluate them or simplify them. We solve equations (typically equalities, like $x = 2x + 4$, but sometimes inequalities, like $x < 3x + 1$). $\endgroup$ Oct 24, 2015 at 1:40
  • $\begingroup$ "Simplify" is not a well-defined notion, I prefer $-(\sqrt{x}+\sqrt{x+1})^2$ over the multiplied out version. But your tracher may not.. $\endgroup$ Oct 24, 2015 at 1:40
  • $\begingroup$ @JohnHughes Thank you $\endgroup$
    – Mehdi
    Oct 24, 2015 at 1:42

1 Answer 1


You are correct; however, there are a few things that I think are worth noting.

First, $x$ is presumably taken to be a positive real number in the initial expression, and because of this you can say, as you have, that

$$\sqrt{x} \, \sqrt{x+1}=\sqrt{x(x+1)} \; , \;x \in \mathbb{R}^+$$

For any complex number $z \in \mathbb{C}$ this would not hold (e.g. it hold for negative real values). A quick way to see this would be to observe that allowing negative $x$ would imply

\begin{equation} 1=\sqrt{1}=\sqrt{-1 \cdot -1}=\sqrt{-1}\sqrt{-1}=i \cdot i = i^2 = -1 \\ \end{equation}

which is a clear contradiction.

An extension of this note would be to always be aware of what your variable is and note the restrictions that applying certain operations impose. For instance, one would not want to write

$$x=\sqrt{x^2} \; \forall x \in \mathbf{R}$$

The above is simply not true, for suppose $x=-1$, then this says that


which is another clear contradiction.

In short, never get too caught up in routine manipulations that you forget the mathematics involved; always do a quick sanity check.

Second, one does not "solve" an expression. I believe you know this, but it is all too common for algebra students in high school to be very confused by the distinction between functions and their roots, equations, expressions, and so forth. What you are looking to do is find an equivalent expression to the one initially stated, that is, you would like to simplify the expression.

Third, the idea of "simplifying" as far as possible is a nebulous one. There exist many scenarios, infinitely many in fact, where an expression can be simplified in many different ways, at least two of which are equally as "nice". Trivially, we have that if some expression, $y$, simplifies to, say, $ax^n+c$, then it also simplifies to $c+ax^n$, and both are arguably just as good; I call this trivial because addition is commutative so the order does not matter, but there are better examples where the expressions actually differ somewhat substantially in form but are equivalent and can be reasonably considered as being equally aesthetic.

Hopefully, this has helped you in some way or provided you with additional insights into the problem. But, the question is fairly straightforward and you indeed did get the correct answer. I too believe your final simplification is the intended one. So, you did a good job!


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .