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Solve for c: \(\sqrt{4+\sqrt{8+4c}}+ \sqrt{2+\sqrt{2+c}} = 2+2\sqrt{2}\)

 Apr 4, 2020
 #1
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c = 4.

 Apr 4, 2020
 #2
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We split the equations:

(1) \(\sqrt{4+\sqrt{8+4c}}=2\)

(2) \(\sqrt{2+\sqrt{2+c}}=2\sqrt{2}\)

 

Square both sides of both equations:

(1) \(4+\sqrt{8+4c}=4\)

(2) \(2+\sqrt{2+c}=8\)

 

Simplify and square both sides of both equations once again:

(1) \(8+4c=0\)

(2) \(2+c=36\)

 

We get:

\(c=-2\) and \(c=34\)

 

That is incorrect, because \(c\) needs to be the same value for the equations to work.

By commutative property, we get another two equations:

(3) \(\sqrt{4+\sqrt{8+4c}}=2\sqrt{2}\)

(4) \(\sqrt{2+\sqrt{2+c}}=2\)

 

Square both sides of both equations:

(3) \(4+\sqrt{8+4c}=8\)

(4) \(2+\sqrt{2+c}=4\)

 

Simplify and sqaure both sides of both equations once again:

(3) \(8+4c=16\)

(4) \(2+c=4\)

 

For both equations, we get \(\boxed{c=2}\)

 

Take that! Guest!

 Apr 5, 2020

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