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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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10 Apr 2019, 01:12
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Difficulty:

45% (medium)

Question Stats:

72% (01:44) correct 28% (01:51) wrong based on 25 sessions

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If $$p = \frac{\sqrt{3} - 2}{\sqrt{2} +1}$$, then which one of the following equals p – 4?

(A) $$\sqrt{3} – 2$$

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

(C) 2

(D) $$-2 \sqrt{2} + \sqrt{6} – \sqrt{3} – 2$$

(E) $$-2 \sqrt{2} + 6 – \sqrt{3} + 2$$

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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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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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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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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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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

Posted from my mobile device
Manager
Joined: 19 Apr 2017
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Concentration: General Management, Sustainability
Schools: ESSEC '22
GPA: 3.9
WE: Operations (Hospitality and Tourism)
If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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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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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 )

Intern
Joined: 15 Feb 2019
Posts: 13
Re: If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following  [#permalink]

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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 )

Wow! Thank you for stating this! It's stuck in my head now! Thanks!!
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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