Any way to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}$?

Question

I was solving a radical equation $x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} = 2$. I deduced it to $\sqrt{x } + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}.$ Answer is $\frac1{24}$.

The first equation has two solutions however the latter one has only one solution. I lossed up one solution but still, I'm interested in solving it.

I wish if someone could help me in solving any of the above equations.

Here's my work:

$$\begin{align}x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} &= 2\\x + \frac{1}{2}\left(2 \sqrt{x}\sqrt{x+1} + 2\sqrt{x+1}\sqrt{x+2} + 2\sqrt{x}\sqrt{x+2}\right) &=2\tag{1}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (x + x + 1 + x + 2)\Big]& = 2\tag{2}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 2\\2x + \Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 4\\(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 & = x + 7\\\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} & = \sqrt{x+7}\end{align}$$ Moving from $(1)$ to $(2)$, I used $2(ab + bc + ca) = (a+b+c)^2 - (a^2+b^2+c^2)$.

---

In general, is it possible to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+k}$ by hand?

Answer 1

Given the equation $$ \sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7} $$ consider it in the form $$ \sqrt{x} + \sqrt{x+1} = \sqrt{x+7} - \sqrt{x+2} $$ then by squaring both sides the result is $$ \sqrt{x (x+1)} = 4 - \sqrt{(x+2)(x+7)}. $$ Now, with this equation written as $4 = \sqrt{x (x+1)} + \sqrt{(x+2)(x+7)}$ then, by squaring both sides, $$ \sqrt{x (x+1) (x+2) (x+7)} = x^2 + 5 x -1. $$ Again squaring both sides leads to $$ x (x+1) (x+2)(x+7) = x^4 + 10 x^3 + 23 x^2 - 10 x + 1 = x(x+1)(x+2)(x+7) - 24 x + 1$$ which gives the equation $$24 x = 1 \hspace{5mm} \text{and the result} \, x = \frac{1}{24}. $$

Answer 2

There is a fairly simple approach to this:

$\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}=\sqrt{x+7}$

$\sqrt{x}+\sqrt{x+1}=\sqrt{x+7}-\sqrt{x+2}$

Now, squaring both sides, we get:

$2x+1+2\sqrt{x(x+1)}=2x+9-2\sqrt{(x+2)(x+7)}$

$\sqrt{x(x+1)}=4-\sqrt{(x+2)(x+7)}$

Squaring once again:

$x(x+1)=16-8\sqrt{(x+2)(x+7)}+(x+2)(x+7)$

$x^2+x=16-8\sqrt{(x+2)(x+7)}+x^2+9x+14$

$8\sqrt{(x+2)(x+7)}=30+8x$

$4\sqrt{(x+2)(x+7)}=15+4x$

Squaring once again:

$16(x+2)(x+7)=225+120x+16x^2$

$16(x^2+9x+14)=225+120x+16x^2$

From here, the solution is very straight forward with a simple quadratic equation that you can solve easily. Make sure to verify the roots you get by plugging them in and checking if they satisfy the original equation. Some of them could be extraneous due to squaring of both sides.

Answer 3

Let $x+1=y, \;y\geqslant 1$, then you have:

$$ \begin{align}\sqrt y+\sqrt {y-1}+\sqrt {y+1}=\sqrt {y+6}\end{align} $$

Dividing both side of the equation by $\sqrt y$ and using the substitution $\dfrac 1y=u,\;u>0$, then you obtain:

$$ \begin{align}&\sqrt {1-u}+\sqrt {1+u}=\sqrt {1+6u}-1\\ \implies &\sqrt {1-u^2}=3u-\sqrt {1+6u}\\ \implies &10u^2+6u=6u\sqrt {1+6u}\\ \implies &3\sqrt {6u+1}=5u+3\\ \implies &9(6u+1)=(5u+3)^2\\ \implies &u=\frac {24}{25}\\ \implies &x=\frac {1}{24}.\end{align} $$

Answer 4

I’ll go by your original radical equation as it can be easily factorized for $x\ge0$. Since its domain is $(-\infty,-2]\cup[0,\infty)$, it is imminent to break it into $2$ cases so that in each case, we can distribute the square root.

$\textbf{Case 1, $x\ge0$:}$

$$x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} = 2$$ $$\require{cancel}\implies\left(\sqrt{x+1}+\sqrt x\right)\cancel{\left(\sqrt{x+2}+\sqrt x\right)}=\cancel{\left(\sqrt{x+2}+\sqrt x\right)}\left(\sqrt{x+2}-\sqrt x\right)$$ $$\implies 2\sqrt x+\sqrt{x+1}=\sqrt{x+2}$$

Squaring both sides, we get

$$5x+1+4\sqrt{x^2+x}=x+2$$ $$\implies 4\sqrt{x^2+x}=1-4x$$

Squaring both sides again, we get

$$\cancel{16x^2}+16x=\cancel{16x^2}-8x+1$$ $$\implies x=\frac1{24}$$

Note that $\frac1{24}>0$, i.e., it is indeed a solution to the original equation and not an extraneous one.

$\textbf{Case 2, $x\le-2$:}$

To avoid any confusions in signs, it’s useful to substitute $t=-x(\ge2)$:

$$-t+\sqrt{t(t-1)}+\sqrt{(t-1)(t-2)}+\sqrt{t(t-2)}=2$$ $$\implies 2\sqrt{t(t-1)}+2\sqrt{(t-1)(t-2)}+2\sqrt{t(t-2)} =2t+4$$ $$\implies \left(\sqrt t+\sqrt{t-1}+\sqrt{t-2}\right)^2=5t+1$$ $$\implies \sqrt t+\sqrt{t-1}+\sqrt{t-2}=\sqrt{5t+1}$$

To simplify this equation into a polynomial equation, we’d have to square both sides $3$ times as after the first squaring, there’d be $2$ terms in the square root on both the sides which initiates the need to square a second time so that there is only one square root term. There are $3$ possibilities for rearrange the above equation for squaring, namely

$(1) $$\sqrt t+\sqrt{t-1}=\sqrt{5t+1}-\sqrt{t-2}$$ $

$(2) $$\sqrt t+\sqrt{t-2}=\sqrt{5t+1}-\sqrt{t-1}$$ $

$(3) $$\sqrt{t-1}+\sqrt{t-2}=\sqrt{5t+1}-\sqrt t$$ $

Possibility $(1)$ would be the most convenient one since $0+(-1)=1-2$, so on the first squaring, there would be only a term in $t$ left besides the square roots, which is the easiest to square once again as there is no constant term.

$$\sqrt t+\sqrt{t-1}=\sqrt{5t+1}-\sqrt{t-2}$$ $$\implies 2t\cancel{-1}+2\sqrt{t(t-1)}=6t\cancel{-1}-2\sqrt{(t-2)(5t+1)}$$ $$\implies 2t=\sqrt{t(t-1)}+\sqrt{(t-2)(5t+1)}$$ $$\implies -t^2+5t+1=\sqrt{t(t-1)(t-2)(5t+1)}$$ $$\implies 4t^4-4t^3-16t^2-8t-1=0$$

Wolfram Alpha gives a really messy solution in terms of nested radicals and complex numbers, the solution being $x\approx-2.7391$.

P.S. if the original equation contained $|x|$ instead of $x$, the equation can be factorized easily for $x\le-2$ as well. But then, no solution would exist for $x\le-2$. I leave this exercise to you.

On simplification, we’d be left with $$\sqrt{-x-1}+\sqrt{-x-2}=0$$

Answer 5

We have $x\ge0$. Dividing by the right hand side we have $$\sqrt{\frac{x}{x+7}}+\sqrt{\frac{x+1}{x+7}}+\sqrt{\frac{x+2}{x+7}}=1.$$ Note that the left is monotonic, so there is at most one solution. Further note that $1/24$ is a solution. Therefore it is the only solution.