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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following

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Joined: 02 Sep 2009
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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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New post 10 Apr 2019, 01:12
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

74% (01:51) correct 26% (02:11) wrong based on 34 sessions

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Re: If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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New post 10 Apr 2019, 02:37
Rationalized the equation and subtract 4 from it.

Will give you D

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Re: If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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New post 10 Apr 2019, 03:37
We will rationalize the equation and subtract 4 from it.

IMO D
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Re: If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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New post 10 Apr 2019, 04:21
Formula to remember for this question \(= a^2-b^2 = (a+b)(a-b)\)

\(\frac{√3−2}{√2+1}*\frac{√2-1}{√2-1}=\frac{(√3−2)(√2-1)}{2-1}=(√3−2)(√2-1)\)

\((√3−2)(√2-1) =√2√3-√3-2√2+2=P\)

\(P-4 = √2√3-√3-2√2+2-4 = √6-√3-2√2-2 =−2√2+√6–√3–2\) (with a little rearranging you get D)
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Re: If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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New post 10 Apr 2019, 04:34
AvidDreamer09 wrote:
Formula to remember for this question \(= a^2-b^2 = (a+b)(a-b)\)

\(\frac{√3−2}{√2+1}*\frac{√2-1}{√2-1}=\frac{(√3−2)(√2-1)}{2-1}=(√3−2)(√2-1)\)

\((√3−2)(√2-1) =√2√3-√3-2√2+2=P\)

\(P-4 = √2√3-√3-2√2+2-4 = √6-√3-2√2-2 =−2√2+√6–√3–2\) (with a little rearranging you get D)


WHERE DID YOU GET
√2−1√2−1 From??? Confused

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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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New post 10 Apr 2019, 04:43
1
Peachie

Peachie wrote:
AvidDreamer09 wrote:
Formula to remember for this question \(= a^2-b^2 = (a+b)(a-b)\)

\(\frac{√3−2}{√2+1}*\frac{√2-1}{√2-1}=\frac{(√3−2)(√2-1)}{2-1}=(√3−2)(√2-1)\)

\((√3−2)(√2-1) =√2√3-√3-2√2+2=P\)

\(P-4 = √2√3-√3-2√2+2-4 = √6-√3-2√2-2 =−2√2+√6–√3–2\) (with a little rearranging you get D)


WHERE DID YOU GET
√2−1√2−1 From??? Confused

Posted from my mobile device



So what i did was multiplied (√2−1) to the numerator and denominator.... this does not change the equation but it helps to simplify the equation...

As a rule of thumb try eliminating the denominator when you get questions like these
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Re: If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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New post 10 Apr 2019, 05:30
1
Peachie wrote:
AvidDreamer09 wrote:
Formula to remember for this question \(= a^2-b^2 = (a+b)(a-b)\)

\(\frac{√3−2}{√2+1}*\frac{√2-1}{√2-1}=\frac{(√3−2)(√2-1)}{2-1}=(√3−2)(√2-1)\)

\((√3−2)(√2-1) =√2√3-√3-2√2+2=P\)

\(P-4 = √2√3-√3-2√2+2-4 = √6-√3-2√2-2 =−2√2+√6–√3–2\) (with a little rearranging you get D)


WHERE DID YOU GET
√2−1√2−1 From??? Confused

Posted from my mobile device



We call it rationalization

if u have (√2 -1)/(√2+1)

them u can always multiply the Numerator and denominator by √2-1 ( same term but change the sign of one term )


award kudos if helpful
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Re: If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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New post 10 Apr 2019, 07:40
m1033512 wrote:
Peachie wrote:
AvidDreamer09 wrote:
Formula to remember for this question \(= a^2-b^2 = (a+b)(a-b)\)

\(\frac{√3−2}{√2+1}*\frac{√2-1}{√2-1}=\frac{(√3−2)(√2-1)}{2-1}=(√3−2)(√2-1)\)

\((√3−2)(√2-1) =√2√3-√3-2√2+2=P\)

\(P-4 = √2√3-√3-2√2+2-4 = √6-√3-2√2-2 =−2√2+√6–√3–2\) (with a little rearranging you get D)


WHERE DID YOU GET
√2−1√2−1 From??? Confused

Posted from my mobile device



We call it rationalization

if u have (√2 -1)/(√2+1)

them u can always multiply the Numerator and denominator by √2-1 ( same term but change the sign of one term )


award kudos if helpful


Wow! Thank you for stating this! It's stuck in my head now! Thanks!!
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Re: If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following   [#permalink] 10 Apr 2019, 07:40
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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following

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