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If x=√{4-√(4+...)}, then x=√{4-√(4+x)}.

From the equation above, x is obviously a positive that must not be greater than 2. Options (B),(C),(E) all show that the value of x is greater than 2, contradicting the logic. Thus, confidently eliminate options (B),(C), and (E).

x = √{4-√(4+x)}
x^2 = 4-√(4+x)
√(4+x) = 4-x^2
4+x= (4-x^2)^2
x= (x^2-4)^2 -4

TRY (A) x = (-1+√13)/2
--> x = ((-1+√13)^2/4 -4)^2 -4
--> x = {-(1+√13)/2}^2 -4
--> x = (1+√13)^2/4 -4
--> x = (14+2√13)/4 -4
--> x = (-1+√13)/2 (CORRECT)
Since (A) is correct, we don't have to check (D). For exercise, you can try to validate (D)

FINAL ANSWER IS (A)

Posted from my mobile device

√(4-√(4+√4…))=√(4-√(4+2…))=√(4-√(6…))
2<√6<3~2.4;
4-2.4=1.6;
1<√1.6<1.41(√2)

(B,C,D)>2
(A) 3<√13<4…~3.5-1=2.5/2=1.25
(E) 3.5<√15<4…~3.8-1=2.8/2=1.4

Ans (A)

What is the value of \(\sqrt[]{4-\sqrt[]{4+\sqrt[]{4-\sqrt[]{4+...}}}}\) ?

The value of \(\sqrt[]{4-\sqrt[]{4+\sqrt[]{4-\sqrt[]{4+...}}}}\) is definitely less than 2.

In answer choices, Only the answer choices A and D can satisfy that expression.

--> Let's say that \(\sqrt[]{4-\sqrt[]{4+\sqrt[]{4-\sqrt[]{4+...}}}}\)= \(x\)

--> square both sides:
\(4- \sqrt[]{4+\sqrt[]{4-\sqrt[]{4+...}}}\) = \(x^{2}\)

\(4- x^{2}= \sqrt[]{4+\sqrt[]{4-\sqrt[]{4+...}}}\)

Square both sides again:
-->\((4- x^{2})^{2}= 4+ \sqrt[]{4-\sqrt[]{4+\sqrt[]{4-\sqrt[]{4+...}}}}\)

\( (4- x^{2})^{2}= 4+x\)

Let's check the value of answer choice A (\(\frac{√13-1}{2}\))

\((4- (\frac{√13-1}{2})^{2})^{2}= 4+\frac{√13-1}{2}\)

\((4 -\frac{(14-2√13)}{4})^{2}= \frac{(7+ √13)}{2}\)

\((2+2√13)^{2}= 56+8√13\)

\(4+ 8√13 + 4*13 = 56+8√13\)

\(56+8√13= 56+8√13\)
(CORRECT)

The answer is A.

let the expression be x

after squaring,

x^2 = 4-sqrt(4+x)
x^2-4 = sqrt(4+x)
(x^2-4 )^2 = (4+x)

the value of x must be <2
from options we can eliminate B,C,E as all these values are > 2

so we are left with A,D

after substituting both values in the above equation.

A will be satisfying the equation

OA:A