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.
 A: 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.
A: 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.
A: 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
A: 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.
A: 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 $x$,
$$x+4=4 \implies x=0.$$
A: It's easy to guess that the answer is $x=0$. 
However, one should prove that this is the only root. Here is how.


*

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

*Function $y =1$ is constant.

*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.
A: 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.
