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

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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 $x$, $$x+4=4 \implies x=0.$$