+178 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)
+26393 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:
\(\begin{array}{rcll} \sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}} &\overset{?}{=}& 6 \qquad a = 8\\ \sqrt{4+\sqrt{16+16\cdot 8}}+\sqrt{1+\sqrt{1+8}} & \overset{?}{=} & 6 \\ \sqrt{4+\sqrt{144}}+\sqrt{1+3} & \overset{?}{=} & 6 \\ \sqrt{4+12}+\sqrt{4} & \overset{?}{=} & 6 \\ \sqrt{16}+2 & \overset{?}{=} & 6 \\ 4+2 & \overset{?}{=} & 6 \\ 6 & \overset{!}{=} & 6 \quad \checkmark\\ \end{array}\)
