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(1+ (3)^1/3) (2+(3^1/2))^1/2?

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(1+ (3)^1/3) (2+(3^1/2))^1/2?  [#permalink]

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New post Updated on: 03 Oct 2017, 02:16
12
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

49% (02:26) correct 51% (02:47) wrong based on 161 sessions

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\((1+ \sqrt{3})*(\sqrt{2+\sqrt{3}})\)


A. \(\sqrt{2}(2+\sqrt{3})\)

B. \(\sqrt{2}(3+\sqrt{3})\)

C. \(\sqrt{2}(2-\sqrt{3})\)

D. \(\sqrt{3}(2+\sqrt{3})\)

E. \(\sqrt{2}(2+\sqrt{2})\)

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Originally posted by roastedchips on 06 Jul 2017, 07:18.
Last edited by Bunuel on 03 Oct 2017, 02:16, edited 2 times in total.
Formatted the question.
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Re: (1+ (3)^1/3) (2+(3^1/2))^1/2?  [#permalink]

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New post 06 Jul 2017, 08:20
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Let X = \((1+ \sqrt{3})\sqrt{(2+3^{1/2})}\) = \((1+ \sqrt{3})\sqrt{(1 + 1 + 3^{1/2})}\)

If a = \(1+ \sqrt{3}\), then \(a^2 = 1 + 3 + 2\sqrt{3} = 4 + 2\sqrt{3}\)

The expression now becomes X = \(a * \sqrt{1 + a}\)

Squaring
\(X^2 = a^2(1+a)\)

Substituting values,
\(X^2 = (4 + 2\sqrt{3})(2 + \sqrt{3})\)
\(= 2(2 + \sqrt{3})(2 + \sqrt{3})\)

Taking square root
X = \(\sqrt{2}(2 + \sqrt{3})\) (Option A)
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Re: (1+ (3)^1/3) (2+(3^1/2))^1/2?  [#permalink]

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New post 10 Jul 2017, 09:29
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Easy method,
Let (1+√3)√(2+√3) = x. Square both sides
x^2 = (1+√3)^2 * (2+√3)
= (1 + 3 + 2√3) * (2 + √3)
= (4 + 2√3) * (2 + √3)
= 2(2 + √3) * (2 + √3)
= 2 * (2 + √3)^2.
Take square root of both sides, x = √2 * (2 + √3)
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Re: (1+ (3)^1/3) (2+(3^1/2))^1/2?  [#permalink]

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New post 27 Sep 2017, 09:10
pushpitkc wrote:
Let X = \((1+ \sqrt{3})\sqrt{(2+3^{1/2})}\) = \((1+ \sqrt{3})\sqrt{(1 + 1 + 3^{1/2})}\)

If a = \(1+ \sqrt{3}\), then \(a^2 = 1 + 3 + 2\sqrt{3} = 4 + 2\sqrt{3}\)

The expression now becomes X = \(a * \sqrt{1 + a}\)

Squaring
\(X^2 = a^2(1+a)\)

Substituting values,
\(X^2 = (4 + 2\sqrt{3})(2 + \sqrt{3})\)
\(= 2(2 + \sqrt{3})(2 + \sqrt{3})\)

Taking square root
X = \(\sqrt{2}(2 + \sqrt{3})\) (Option A)



Are we allowed to take 2 common from one bracket set?

eg 6X7=2(3)(7)
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Re: (1+ (3)^1/3) (2+(3^1/2))^1/2?  [#permalink]

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New post 03 Oct 2017, 00:16
gps5441 wrote:
pushpitkc wrote:
Let X = \((1+ \sqrt{3})\sqrt{(2+3^{1/2})}\) = \((1+ \sqrt{3})\sqrt{(1 + 1 + 3^{1/2})}\)

If a = \(1+ \sqrt{3}\), then \(a^2 = 1 + 3 + 2\sqrt{3} = 4 + 2\sqrt{3}\)

The expression now becomes X = \(a * \sqrt{1 + a}\)

Squaring
\(X^2 = a^2(1+a)\)

Substituting values,
\(X^2 = (4 + 2\sqrt{3})(2 + \sqrt{3})\)
\(= 2(2 + \sqrt{3})(2 + \sqrt{3})\)

Taking square root
X = \(\sqrt{2}(2 + \sqrt{3})\) (Option A)



Are we allowed to take 2 common from one bracket set?

eg 6X7=2(3)(7)


Yes you can take '2' common from one braket.

Notice that when you multiply '2' back to the braket, you'll get the previous equation.

I'll explain with example.

Case 1: Let us say an equation is x = (2y + 4)*(y^2 + 3y); here x and y are some variables.
In this case you can take '2' common from first bracket and 'y' common from second bracket. Equation can be written as, x = 2y*(y+2)*(y+3).
You can also multiply '2' and 'y' separately to the brackets, even to the bracket from which each was not taken.
Equation can also be written as, x = (y^2+2y)*(2y+6) = 2*(y^2+2y)*(y+3) = y*(y+2)*(2y+6). All these equations are same and correct.

Case 2: Let us say an equation is x = (2y + 4) + (y^2 + 3y); here x and y are some variables.
In this case also you can take '2' common from first bracket and 'y' common from second bracket but you cannot multiply both commons like in previous case.
Equation will be written as, x = 2(y+2) + y(y+3). This equation is same as original equation.

Case 3: Let us say an equation is x = (3y + 6) + (y^2 + 2y); here x and y are some variables.
In this case also you can take '2' common from first bracket and 'y' common from second bracket. Equation can be written as, x = 3(y+2)+y(y+2).
In this case, you can take '(y+2)' as common and equation will become, x = (y+2)*(y+3).

I hope this clears your doubt.
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Re: (1+ (3)^1/3) (2+(3^1/2))^1/2?  [#permalink]

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New post 03 Oct 2017, 02:15
Very informative thread.
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Re: (1+ (3)^1/3) (2+(3^1/2))^1/2? &nbs [#permalink] 03 Oct 2017, 02:15
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