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

 Aug 18, 2017
 #1
avatar+178 
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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)

 Aug 18, 2017
edited by Jeffes02  Aug 18, 2017
 #2
avatar+22188 
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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:

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

 

laugh

 Aug 18, 2017

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