What is a: sqrt(4+sqrt(16+16a))+sqrt(1+sqrt(1+a))=6.

Input: What is a:   sqrt(4+sqrt(16+16a))+sqrt(1+sqrt(1+a))=6

Intepretation: Solve for $a$ in $\sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}}=6$

Simplify:

$\sqrt{4+4\sqrt{1+a}}+\sqrt{1+\sqrt{1+a}}=6$

We know that $\sqrt{4+4\sqrt{1+a}}$ is $\sqrt4=2$ times larger than $\sqrt{1+\sqrt{1+a}}$

Merge:

$3\sqrt{1+\sqrt{1+a}}=6$

Divide both sides by a factor of 3:

$\sqrt{1+\sqrt{1+a}}=2$

Since we know that $\sqrt4=2$

Therefore:

$1+\sqrt{1+a}=4$

$\sqrt{1+a}=3$

Since $\sqrt9=3$

$1+a=9$

$a=8$

Q.E.D.

(For one to solve this question, you just need to know the basic ideas of squares and powers (And some work) :P)

What is a:   sqrt(4+sqrt(16+16a))+sqrt(1+sqrt(1+a))=6

($\sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}}=6.$)

$\begin{array}{|rcll|} \hline \sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ \sqrt{4+\sqrt{16(1+a)}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ \sqrt{4+4\sqrt{1+a}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ \sqrt{4(1+\sqrt{1+a})}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ 2\sqrt{1+\sqrt{1+a}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ 3\sqrt{1+\sqrt{1+a}} &=& 6 \quad & | \quad : 3 \\ \sqrt{1+\sqrt{1+a}} &=& 2 \quad & | \quad \text{square both sides} \\ 1+\sqrt{1+a} &=& 4 \quad & | \quad -1 \\ \sqrt{1+a} &=& 3 \quad & | \quad \text{square both sides} \\ 1+a &=& 9 \quad & | \quad -1 \\ \mathbf{ a } & \mathbf{=} & \mathbf{8} \\ \hline \end{array}$

Proof:

$\begin{array}{rcll} \sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}} &\overset{?}{=}& 6 \qquad a = 8\\ \sqrt{4+\sqrt{16+16\cdot 8}}+\sqrt{1+\sqrt{1+8}} & \overset{?}{=} & 6 \\ \sqrt{4+\sqrt{144}}+\sqrt{1+3} & \overset{?}{=} & 6 \\ \sqrt{4+12}+\sqrt{4} & \overset{?}{=} & 6 \\ \sqrt{16}+2 & \overset{?}{=} & 6 \\ 4+2 & \overset{?}{=} & 6 \\ 6 & \overset{!}{=} & 6 \quad \checkmark\\ \end{array}$