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

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

Guest Aug 18, 2017
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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)

Jeffes02  Aug 18, 2017
edited by Jeffes02  Aug 18, 2017
#2
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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:

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

heureka  Aug 18, 2017

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