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If the following expression extends to an infinite number of roots, then what is the value of 4−4+4−4+...‾‾‾‾‾√‾‾‾‾‾‾‾‾‾‾‾√‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√4−4+4−4+... ?

A. (√13−1)/2
B. (√13+1)/2
C. (√11+1)/2
D. (√15−1)/2
E. (√15+1)/2

√(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)
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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.
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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
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Hi Bunuel - is the OA really B for this problem? A seems to be correct to me, and it looks like the results of this very intelligent user base support A.
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Hi Bunuel - is the OA really B for this problem? A seems to be correct to me, and it looks like the results of this very intelligent user base support A.

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The OA is A. Edited. Thank you!
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Call the expression X.

Squaring yields 4-(4+X)^.5, so

X^2= 4-(4+X)^.5. Squaring again and rearranging yields:

X^4-8X^2-X+12=0

Try factoring and allocating an X^2 to each (?) and last elements of each have to multiply to 12, correct ?

(X^2...-4)(X^2...-3)

Using minuses above because seems right to generate a negative 8X^2

So far, X^4-7X^2+12. Need a negative X, for sure, and another -X^2.

The -X^2 has to be a +X and a -X multiplied, but which side do each go ?

Let's just try:

(X^2-X-4)(X^2+X-3), which upon expansion does yield our original expression.

One or both of the above equals 0, so:

X= (1+/- (17)^.5)/2 perhaps?

Well, 1<X<2 by inspecting up top, and this expression >2.5, so it must be:

X^2+X-3=0, or

X= (-1+/- (13)^.5)/2

Since X>0, X=((13)^.5)-1)/2

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