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Bunuel
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I

Raise all the options to the power of 4.

I. 2* 5^2 = 50

II. 3*2^2 = 12

III. 401

II < I < III

C

1 is 16*25= 400
2 is 81*4=324
3 is 401

Liked your style of solution though :) i went the conventional way. Took me a bit more time than this solution :o :-D :-D :-D
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Bunuel
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I

Put all numbers into their most basic square root forms.

I. \(2\sqrt{5}\) = \(\sqrt{4*5} = \sqrt{20}\)

II. \(3\sqrt{2}\)
= \(\sqrt{9*2}\) = \(\sqrt{18}\)

III. \(\sqrt[4]{401}\). This one is a bit more difficult. To take the 4th root is to take the square root twice. The square root of 401 is slightly more than 20, so the 4th root of 401 will be slightly more than \(\sqrt{20}\).

Thus, our ordering from smallest to largest is II,I,III.

Answer: C
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Bunuel
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I

Put all numbers into their most basic square root forms.

I. \(2\sqrt{5}\) = \(\sqrt{4*5} = \sqrt{20}\)

II. \(3\sqrt{2}\)
= \(\sqrt{9*2}\) = \(\sqrt{18}\)

III. \(\sqrt[4]{401}\). This one is a bit more difficult. To take the 4th root is to take the square root twice. The square root of 401 is slightly more than 20, so the 4th root of 401 will be slightly more than \(\sqrt{20}\).

Thus, our ordering from smallest to largest is II,I,III.

Answer: C

For the last step I would suggest to bring \(\sqrt{20}\) to \(2\sqrt{5}\)[/b], which is a little bit smaller than I
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Bunuel
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I

I. \((2\sqrt{5})^2\) = 20 & \(20^2\) = \(400\)
II.\((3\sqrt{2})2\) = 18 & \(18^2\) = \(324\)
III.\((\sqrt[4]{401})^4\) = \(401\)

So, Ascending order of arrangement will be II, I, III , answer will be (C)
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Bunuel
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


Best way to answer such questions is APPROXIMATION

I. \(2\sqrt{5}\) = 2*2.25 = 4.5


II. \(3\sqrt{2}\) = 3*1.41 = 4.2


III. \(\sqrt[4]{401}\) = \(\sqrt{20}\) = 4.49


Answer: Option C
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Bunuel
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I

Best way to answer this is \(a^4 = 400, b^4 = 162, c^4 = 401\).. Now if we have doubt in this try\(a^4 - b^4 = (a-b)(a+b)(a^2+b^2)\).. Similarly,\(c^4 - a^4, c^4 - b^4\)
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