# Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$

Can someone give me some hints on how to start solving $\sqrt{x+4}-\sqrt{x+1}=1$ for x?

Like I tried to factor it expand it, or even multiplying both sides by its conjugate but nothing comes up right.

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$0$ of course, it's easy to guess ... square it, isolate the roots and then square it. – Santosh Linkha May 23 '13 at 17:10
This is pretty much a duplicate of math.stackexchange.com/questions/399199/… – Thomas May 23 '13 at 20:38

Start by squaring it to get

$$x+4-2\sqrt{(x+4)(x+1)}+x+1=1\;,$$

which simplifies to

$$\sqrt{(x+4)(x+1)}=x+2\;.$$

Now square again.

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Thanks for the help! – Regis May 23 '13 at 17:11
@Regis: You’re welcome! – Brian M. Scott May 23 '13 at 17:11
Note - when squaring like this, always check back to see that your solution works for the original equation, since squaring destroys the distinction between the positive and negative values of the square root, and you will generally get solutions corresponding to both. – Mark Bennet May 23 '13 at 17:32

HINT:

As $(x+4)-(x+1)=3 \ \ \ \ \$

$\implies (\sqrt{x+4}-\sqrt{x+1})(\sqrt{x+4}+\sqrt{x+1})=3$

$$\text{As }\sqrt{x+4}-\sqrt{x+1}=1\ \ \ \ \ (1)$$

$$\implies \sqrt{x+4}+\sqrt{x+1}=3\ \ \ \ \ (2)$$

Add/subtract $(1)$ and $(2),$ then square

Generalization :

$$\text{As }(ax+b)-(ax+c)=b-c$$

$$\text{If }\sqrt{ax+b}-\sqrt{ax+c}=d \ \ \ \ \ (1)$$

$$\text{As } (ax+b)-(ax+c)=(\sqrt{ax+b}-\sqrt{ax+c})(\sqrt{ax+b}+\sqrt{ax+c})$$

$$\implies \sqrt{ax+b}+\sqrt{ax+c}=\frac{b-c}d\ \ \ \ \ (2)$$

Add/subtract $(1)$ and $(2),$ then square

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(+1) neat approach – vadim123 May 23 '13 at 17:28
– lab bhattacharjee May 23 '13 at 17:58
very nice approach – srijan May 23 '13 at 18:32

Multiplying by the conjugate as you originally suggested does work here. If you multiply both sides by $\sqrt{x+4} + \sqrt{x+1}$ you get $$(x+4) - (x+1) = \sqrt{x+4} + \sqrt{x+1}$$ Which is the same as $$\sqrt{x+4} + \sqrt{x+1} = 3$$ Add this to the original equation and divide by $2$ to obtain $$\sqrt{x+4} = 2$$ Squaring you get $$x +4 = 4$$ Therefore $x = 0$ is the only solution.

Also note the similarity to lab bhattacharjee's method.

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Such equations, if slick tricks such as lab bhattacharjee's can't apply, are solved with a standard procedure:

\begin{align} &\sqrt{x+4}-\sqrt{x+1}=1\\[2ex] &\text{Rearrange}\\ &\sqrt{x+4}=1+\sqrt{x+1}\\[2ex] &\text{Square}\\ &x+4=1+2\sqrt{x+1}+(x+1)\\[2ex] &\text{Rearrange}\\ &2=2\sqrt{x+1}\\[2ex] &\text{Simplify}\\ &1=\sqrt{x+1}\\[2ex] &\text{Square}\\ &1=x+1\\[2ex] &x=0 \end{align}

We just need to ckeck that the solution makes the square roots existent, because at each "Square" stage we are dealing with non negative numbers. Of course the conditions are $$\begin{cases} x\ge-4\\ x\ge-1 \end{cases}$$ which boil down to $x\ge-1$, that's satisfied by our solution.

Some care has to be reserved in different situations, when there's no guarantee that at the "Square" staged we have non negative numbers.

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Note that lab's trick 'always' applies, in some sense; with an equation of form $\sqrt{x+a}-\sqrt{x}=b$ (obviously the given equation can be put into such a form by using $y=x+1$), then by factoring the left side of $x+a-x=a$ one gets $\sqrt{x+a}+\sqrt{x} = \frac{a}{b}$, and then adding/subtracting gives the answer. – Steven Stadnicki May 23 '13 at 17:47
@StevenStadnicki It doesn't work with $\sqrt{2x+a}-\sqrt{x}=b$, for instance; that's what I was referring to. I believe that a standard procedure should be taught, leaving slick tricks to the best students. We don't want to teach math as a collection of tricks, do we? – egreg May 23 '13 at 17:50

Let $a = \sqrt{x+4}$ and $b=\sqrt{x+1}$. So that $a^2 = x + 4$ and $b^2 = x + 1$.

From the given equation, $$\sqrt{x+4} - \sqrt{x+1}=1 \implies a - b = 1 \ \ \ \text{and} \ \ \ a^2-b^2=3.$$ So we have that, $$a^2-b^2 =(a-b)(a+b)=1 \cdot(a+b)=3 \implies a+b=3.$$ We see that $$(a+b)+(a-b) = 2a.$$ Also, $$(a+b)+(a-b)=3+1=4.$$ Therefore, $$2a=4 \implies a=2 \implies \sqrt{x+4}=2.$$ Solving for $x$, $$x+4=4 \implies x=0.$$

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Very interesting approach! – Gamma Function Jun 12 '13 at 5:30
I made an edit to add $\LaTeX$ and brief explanations. This is your answer and you should edit/modify it as you see fit. – Gamma Function Jun 12 '13 at 5:43

It's easy to guess that the answer is $x=0$. However, one should prove that this is the only root. Here is how.

1. Because $\sqrt {x+4} > \sqrt {x+1}$, this function is strictly increasing.

2. Function $y =1$ is constant.

3. It is easy to proove that monotoniously increasing function and a constant can cross at no more than 1 point (either cross at 1 point or do not cross at all).

Since we have a point of $x = 0$, then there are no other roots for this equasion.

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#4 is not true... $x$ and $x + {1 \over 2} \sin x$ are both monotone increasing but their graphs intersect repeatedly. – Zarrax May 23 '13 at 21:35
@Zarrax, Sorry for that. It is the best illustation of my mistake. I'll edit the answer. – Stoleg May 24 '13 at 13:57

You have two numbers $\sqrt{x+4}$ and $\sqrt{x+1}$. Consider their squares, which are (x+4) and (x+1). So, their squares differ by 3, with the larger square number equaling (x+4). Do we know of two square numbers which differ by 3? As you might recall, the square numbers belong to the sequence (1, 4, ...). So, the larger square number (x+4) equals 4, and the smaller square number (x+1) equals 1. Thus, x=0.

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I have to wonder what the people who use a bunch of algebraic manipulation to solve these sorts of problems think about solutions like these or if my answer would have gotten accepted had I managed to post it 20 hours earlier. – Doug Spoonwood May 24 '13 at 14:48
There's also the possibility that $x$ is not a whole number. There are many pairs of positive real numbers whose squares differ by $3$. – Zarrax May 24 '13 at 16:10
@Zarrax Your second statement "There are many pairs of positive real numbers whose squares differ by 3." is true. However, if there exists a possibility that x is not a whole number, then since x works to solve the equation, then there would exist a possibility that x=y and x=z where y does not equal z. In which case, we have a contradiction. So, unless your logic is inconsistent, x=0 here. – Doug Spoonwood May 24 '13 at 16:27
I have no idea what you're saying. At any rate, if you can cover the general real case I suggest adding it to your answer. – Zarrax May 24 '13 at 19:59