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605-655 (Medium)|   Algebra|      
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Bunuel
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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following 10 Apr 2019, 01:12
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71% (01:57) correct 29% (02:10) wrong based on 96 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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D
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Sanjeetgujrall
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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following 10 Apr 2019, 02:37
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Rationalized the equation and subtract 4 from it.

Will give you D

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leonidas56
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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following 10 Apr 2019, 03:37
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We will rationalize the equation and subtract 4 from it.

IMO D
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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following 10 Apr 2019, 04:21
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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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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following 10 Apr 2019, 04:34
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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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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 10 Apr 2019, 04:43
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Peachie

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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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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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m1033512
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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following 10 Apr 2019, 05:30
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Peachie
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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
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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 )


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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following 10 Apr 2019, 07:40
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m1033512
Peachie
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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
Show more

Wow! Thank you for stating this! It's stuck in my head now! Thanks!!
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If p = (3^(1/2) - 2)/(2^(1/2) + 1), then which one of the following 11 Mar 2024, 02:27
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