Rooted away

Let $a = \sqrt{4-x}, b = \sqrt{5-x}, c = \sqrt{6-x}$ . This gives $x = ab + bc + ca$ . Since $a^2 = 4-x$ , we add $a^2$ to both sides to obtain:

$4 = (4-x)+x = a^2 + ab + bc + ca = (a+b)(a+c).$

Similarly, we have:

$5 = (b+c)(b+a), 6 = (c+a)(c+b).$

Multiplying these three equations together and taking the square root, we get $(a+b)(a+c)(b+c) = \sqrt{120}$ . And dividing this by each of the three equations, we get:

$\begin{aligned} b+c &= \frac{\sqrt{120}} 4, \\ c+a &= \frac{\sqrt{120}} 5,\\ a+b &=\frac{\sqrt{120}}6.\end{aligned}$

Now we add these three equations together and divide by two to obtain $a+b+c = \frac{37}{120}\sqrt{120}$ , then subtract this by the first equation to get

$a = \frac{7}{120}\sqrt{120} \implies x = 4-a^2 = \frac{431}{120}.$

$x = \sqrt{4-x}\sqrt{5-x} + \sqrt{5-x}\sqrt{6-x} + \sqrt{6-x}\sqrt{4-x}$

$x - \sqrt{4-x}\sqrt{5-x} = \sqrt{5-x}\sqrt{6-x} + \sqrt{6-x}\sqrt{4-x}$

$x^2 - 2x\sqrt{4-x}\sqrt{5-x} + (4-x)(5-x) = (5-x)(6-x) + 2(6-x)\sqrt{4-x}\sqrt{5-x} + (6-x)(4-x)$

$2(6-x)\sqrt{4-x}\sqrt{5-x} + 2x\sqrt{4-x}\sqrt{5-x} = x^2+ (4-x)(5-x) - (5-x)(6-x) - (6-x)(4-x)$

$6\sqrt{4-x}\sqrt{5-x} = 6x - 17$

$36(4-x)(5-x) = (6x - 17)^2$

$431x - 120 = 0$

$x = \frac{431}{120} < 4$

Hence $a + b = 551$

First substitute $y=5-x$ .

We can then rewrite the equation as:

$5-y+(y)=\sqrt{y+1}(\sqrt{y-1}+\sqrt{y})+\sqrt{y}(\sqrt{y-1}+\sqrt{y})$ .

Simplify that to get $(\sqrt{y-1}+\sqrt{y})(\sqrt{y+1}+\sqrt{y})=5$ .

Multiply both sides by $(\sqrt{y+1}-\sqrt{y})$ to get:

$(\sqrt{y-1}+\sqrt{y})=5(\sqrt{y+1}-\sqrt{y})$ which is:

$5\sqrt{y+1}-6\sqrt{y}=\sqrt{y-1}$ .

Square both sides and simplify to get $30y+13=30\sqrt{y+1}\sqrt{y}$ .

Square both sides again and simplify to get $169y-120$ so $y=\frac{169}{120}$ . Plug this value back in to check if it is extraneous and it is not. $5-x=y$ so $x=\frac{431}{120}$ which is less than four so it works. $431+120=551$ .

It is easy to see that $0 \le x \le 4$ .

Moreover, $0 < x < 4$ .

Suppose that $x=2+2\cos\theta$ for some $\theta \in (0,\pi)$ .

Therefore,

$2+2\cos\theta = \sqrt{\left(2-2\cos\theta\right)\left(3-2\cos\theta\right)}+\sqrt{\left(3-2\cos\theta\right)\left(4-2\cos\theta\right)}+\sqrt{\left(4-2\cos\theta\right)\left(2-2\cos\theta\right)}.$

Dividing by 4 yields

$\frac{1+\cos\theta}{2} = \sqrt{\left(\frac{1-\cos\theta}{2}\right)\left(\frac{1}{4}+\frac{1-\cos\theta}{2}\right)}+\sqrt{\left(\frac{1}{4}+\frac{1-\cos\theta}{2}\right)\left(\frac{1}{2}+\frac{1-\cos\theta}{2}\right)}+\sqrt{\left(\frac{1}{2}+\frac{1-\cos\theta}{2}\right)\left(\frac{1-\cos\theta}{2}\right)}.$

Use some trigonometric identities and get

$\cos^2\frac{\theta}{2} = \sin\frac{\theta}{2} \sqrt{\frac{1}{4}+\sin^2\frac{\theta}{2}}+\sqrt{\left(\frac{1}{4}+\sin^2\frac{\theta}{2}\right)\left(\frac{1}{2}+\sin^2\frac{\theta}{2}\right)}+\sin\frac{\theta}{2}\sqrt{\frac{1}{2}+\sin^2\frac{\theta}{2}}.$

Since $\sin\frac{\theta}{2} \in \mathbb{R}^{+}$ , suppose that $\sin\frac{\theta}{2}=\frac{\tan\alpha}{2}$ for some $\alpha \in \left(0,\frac{\pi}{2}\right)$ .

Hence,

$1-\frac{\tan^2\alpha}{4} = \frac{\tan\alpha}{2}\cdot\frac{\sec\alpha}{2}+\frac{\sec\alpha}{2}\sqrt{\frac{1}{2}+\frac{\tan^2\alpha}{4}}+\frac{\tan\alpha}{2}\sqrt{\frac{1}{2}+\frac{\tan^2\alpha}{4}}.$

Which is

$4-\tan^2\alpha = \tan\alpha \sec\alpha+(\sec\alpha+\tan\alpha)\sqrt{2+\tan^2\alpha}.$

Replace $4$ by $4(\sec^2\alpha-\tan^2\alpha)$ and get

$4\sec^2\alpha-\sec\alpha\tan\alpha-5\tan^2\alpha = (\sec\alpha+\tan\alpha)\sqrt{2+\tan^2\alpha}.$

Or

$(4\sec\alpha-5\tan\alpha)(\sec\alpha+\tan\alpha) = (\sec\alpha+\tan\alpha)\sqrt{2+\tan^2\alpha}.$

Since $\alpha \in \left(0,\frac{\pi}{2}\right)$ , $\sec\alpha+\tan\alpha >0$ . Thus,

$4\sec\alpha-5\tan\alpha = \sqrt{2+\tan^2\alpha}.$

Squaring implies

$16\sec^2\alpha-40\sec\alpha\tan\alpha+25\tan^2\alpha = 2+\tan^2\alpha.$

Replace $2$ by $2(\sec^2\alpha-\tan^2\alpha)$ and get

$14\sec^2\alpha-40\sec\alpha\tan\alpha+26\tan^2\alpha = 0.$

Which is

$7\sec^2\alpha-20\sec\alpha\tan\alpha+13\tan^2\alpha = 0.$

Multiplying $\cos^2\alpha$ and get

$13\sin^2\alpha-20\sin\alpha+7=0.$

Or

$(13\sin\alpha-7)(\sin\alpha-1)=0.$

Since $\alpha \in \left(0,\frac{\pi}{2}\right)$ , $\sin\alpha-1 <0$ . Hence,

$13\sin\alpha-7=0.$

Which is $\sin\alpha=\frac{7}{13}$ .

Therefore $\tan\alpha=\frac{7}{\sqrt{120}}$ .

Then $\sin \frac{\theta}{2} = \frac{\tan\alpha}{2} = \frac{7}{\sqrt{480}}$ .

Consequently, $\cos\theta=1-2\sin^2\frac{\theta}{2}=\frac{191}{240}$ .

Finally, $x=2+2\cos\theta=\frac{431}{120}$ .

Since $\gcd(431,120)=1$ , the answer of the equation is $431+120=551$ .

by caculator

We have $x=\sqrt{4-x}\sqrt{5-x}+\sqrt{5-x}\sqrt{6-x}+\sqrt{6-x}\sqrt{4-x}$ , reorder the equation to obtain: $x=\sqrt{5-x}(\sqrt{4-x}+\sqrt{6-x})+\sqrt{4-x}\sqrt{6-x}$ .

Now, let $t=\sqrt{4-x}\sqrt{6-x}$ to simplify and write less:

$x=\sqrt{5-x}(\sqrt{4-x}+\sqrt{6-x})+t$

Let's perform some algebra:

$x-t=\sqrt{5-x}(\sqrt{4-x}+\sqrt{6-x})$

$(x-t)^{2}=(5-x)(\sqrt{4-x}+\sqrt{6-x})^{2}$

$x^{2}-2xt+t^{2}=(5-x)(4-x+2t+6-x)$ (observe that the term $t$ appeared again)

$x^{2}-2xt+t^{2}=(5-x)(10-2x+2t)$

$x^{2}-2xt+t^{2}=50-10x+10t-10x+2x^{2}-2xt$

Simplify the term $t^{2}$ and cancel some terms:

$x^{2}+(4-x)(6-x)=50-10x+10t-10x+2x^{2}$

$x^{2}+24-10x+x^{2}=50-10x+10t-10x+2x^{2}$

$2x^{2}+24-10x=50-10x+10t-10x+2x^{2}$

$24=50-10x+10t$

$5t=5x-13$

$25t^{2}=(5x-13)^{2}$

Simplify again the term $t^{2}$ :

$25(4-x)(6-x)=25x^{2}-130x+169$

$25x^{2}-250x+600=25x^{2}-130x+169$

$600-169=250x-130x$

$431=120x$

$x=\frac{431}{120}$

Hence, $a=431$ , $b=120$ and $a+b=\boxed{551}$ .

If we solve the equation, we get:

$x=\sqrt{4-x}\sqrt{5-x}+\sqrt{5-x}\sqrt{6-x}+\sqrt{6-x}\sqrt{4-x}$

$x=\frac{431}{120}$

So: $a=431, b=120$ .

Thus, the answer is: $a+b = 431+120 = \boxed{551}$