Another surd

$\sqrt{11+\sqrt{21}} = \sqrt{\left(\sqrt{\dfrac{21}{2}}\right)^2+\left(\sqrt{\dfrac{1}{2}}\right)^2 + 2\sqrt{\dfrac{21}{2}}\sqrt{\dfrac{1}{2}} }=\sqrt{\left(\sqrt{\dfrac{21}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt{\dfrac{21}{2}}+\sqrt{\dfrac{1}{2}}$

Thus $a+b+c+d=21+2+1+2=\boxed{26}$

$\sqrt{11+\sqrt{21}}^{2}=\frac{a}{b}+\frac{c}{d}+2\sqrt{\frac{ac}{bd}}$

$\frac{ad+bc}{bd}=11$

$4\frac{ac}{bd}=21$

From the last equation we see that bd must be a multiple of 4. Since we can choose 2 variables free we take b=1 and d=4 following:

$ac=21$

$a+\frac{c}{4}=11$

Calculating a=1/2 and c=42 we may write: $\sqrt{\frac{1}{2}}+\sqrt{\frac{42}{4}}=\sqrt{\frac{1}{2}}+\sqrt{\frac{21}{2}}$ resuting in the solution $\boxed{26}$ .

Yes I do such problem just as per Challenge Master note for Nihar Mahajan .
$X+Y=11,\ \ X*Y=\dfrac {\sqrt{21}} 2.\ \ \ \ So\ X=\dfrac {21} 2, \ \ Y=\dfrac 1 {2}.$

a+b+c+d=21+2+1+2=26.

We want to express $\sqrt{11+\sqrt{21}}$ as $\sqrt{\dfrac{a}{b}} + \sqrt{\dfrac{c}{d}}$

Now, let $\sqrt{\dfrac{a}{b}} = p$ and $\sqrt{\dfrac{c}{d}} =q$

$\sqrt{11+\sqrt{21}} = p + q\\ 11+\sqrt{21} = (p + q)^2\\ 11+\sqrt{21} = p^2 + q^2 + 2pq$

We know that $p^2 + q^2$ do not have a square root anymore, therefore

$p^2+q^2 = 11$

which leaves us with

$2pq = \sqrt{21}\\ 4p^2q^2 = 21\\ q^2 = \dfrac{21}{4p^2}$

Substitute this to the equation above:

$p^2 + \dfrac{21}{4p^2} = 11\\ 4p^4 - 44 p^2 + 21 = 0\\ (2p^2 -21)(2p^2 - 1) = 0\\ p^2 = \dfrac{21}{2}, \; \dfrac{1}{2}\\ p = \pm \sqrt{\dfrac{21}{2}}, \; \pm \sqrt{\dfrac{1}{2}}$

Now, we know that $p = \sqrt{\dfrac{a}{b}}$ can only either be a positive number or a complex number, there are no negative values for $p$ .

Therefore, $p= \sqrt{\dfrac{21}{2}}, \; \sqrt{\dfrac{1}{2}}$

If $p = \sqrt{\dfrac{21}{2}}, \; q=\sqrt{\dfrac{1}{2}}$ (Similarly, $q$ cannot have a negative value)

If $p = \sqrt{\dfrac{1}{2}}, \; q=\sqrt{\dfrac{21}{2}}$

Therefore, $\sqrt{11+\sqrt{21}} = p + q = \sqrt{\dfrac{1}{2}} + \sqrt{\dfrac{21}{2}}$

$a=1, b=2, c=21, d=2$ . (Or if you simpy insist, $a=21, b=2, c=1, d=2$ )

$a+b+c+d = 1 + 2 + 21 + 2 = \boxed{26}$

Applying the formula $\sqrt{a + \sqrt{b}} = \sqrt{\dfrac{a + \sqrt{a^2 - b}}{2}} + \sqrt{\dfrac{a - \sqrt{a^2 - b}}{2}}$ , we get:-

$\begin{aligned} \sqrt{11 + \sqrt{21}} & = \sqrt{\dfrac{11 + \sqrt{{11}^2 - 21}}{2}} + \sqrt{\dfrac{11 - \sqrt{{11}^2 - 21}}{2}}\\ \\ & = \sqrt{\dfrac{11 + \sqrt{121 - 21}}{2}} + \sqrt{\dfrac{11 - \sqrt{121 - 21}}{2}}\\ \\ & = \sqrt{\dfrac{11 + \sqrt{100}}{2}} + \sqrt{\dfrac{11 - \sqrt{100}}{2}}\\ \\ & = \sqrt{\dfrac{11 + 10}{2}} + \sqrt{\dfrac{11 - 10}{2}}\\ \\ & = \sqrt{\dfrac{21}{2}} + \sqrt{\dfrac{1}{2}} \end{aligned}$

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$\therefore$ $a + b + c + d = 21 + 2 + 1 + 2 = \boxed{26}$