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1/(2-√3)^2=?

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1/(2-√3)^2=?  [#permalink]

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New post 21 Sep 2018, 01:26
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[Math Revolution GMAT math practice question]

\(\frac{1}{(2-√3)^2}=?\)

\(A. 2+√3\)
\(B. 2-√3\)
\(C. 7+4√3\)
\(D. 7-4√3\)
\(E. 4+7√3\)

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Re: 1/(2-√3)^2=?  [#permalink]

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New post 21 Sep 2018, 01:34
1
1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

\(\frac{1}{(2-√3)^2}=?\)

\(A. 2+√3\)
\(B. 2-√3\)
\(C. 7+4√3\)
\(D. 7-4√3\)
\(E. 4+7√3\)


\(\frac{1}{(2-√3)^2}=\frac{1}{(4+3-2*2√3)}=\frac{1}{(7-4√3)}\)

Now multiplying \((7+4√3)\) in both numerator and denominator (called rationalization) we get,

\(\frac{1}{(2-√3)^2}=\frac{1}{(4+3-2*2√3)}=\frac{1}{(7-4√3)}=\frac{(7+4√3)}{(7+4√3)(7-4√3)} = \frac{(7+4√3)}{(7^2-4^2*3)} = (7+4√3)\)

Answer: Option C
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Re: 1/(2-√3)^2=?  [#permalink]

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New post 21 Sep 2018, 01:57
GMATinsight wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

\(\frac{1}{(2-√3)^2}=?\)

\(A. 2+√3\)
\(B. 2-√3\)
\(C. 7+4√3\)
\(D. 7-4√3\)
\(E. 4+7√3\)


\(\frac{1}{(2-√3)^2}=\frac{1}{(4+3-2*2√3)}=\frac{1}{(7-4√3)}\)

Now multiplying \((7+4√3)\) in both numerator and denominator (called rationalization) we get,

\(\frac{1}{(2-√3)^2}=\frac{1}{(4+3-2*2√3)}=\frac{1}{(7-4√3)}=\frac{(7+4√3)}{(7+4√3)(7-4√3)} = \frac{(7+4√3)}{(7^2-4^2*3)} = (7+4√3)\)

Answer: Option C


Can you explain how you got 7 + 4√3?
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Re: 1/(2-√3)^2=?  [#permalink]

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New post 21 Sep 2018, 02:04
1
ArjunJag1328 wrote:
GMATinsight wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

\(\frac{1}{(2-√3)^2}=?\)

\(A. 2+√3\)
\(B. 2-√3\)
\(C. 7+4√3\)
\(D. 7-4√3\)
\(E. 4+7√3\)


\(\frac{1}{(2-√3)^2}=\frac{1}{(4+3-2*2√3)}=\frac{1}{(7-4√3)}\)

Now multiplying \((7+4√3)\) in both numerator and denominator (called rationalization) we get,

\(\frac{1}{(2-√3)^2}=\frac{1}{(4+3-2*2√3)}=\frac{1}{(7-4√3)}=\frac{(7+4√3)}{(7+4√3)(7-4√3)} = \frac{(7+4√3)}{(7^2-4^2*3)} = (7+4√3)\)

Answer: Option C


Can you explain how you got 7 + 4√3?


To rationalize and fracton, we need to multiply the conjugate

Conjugate of a+b is a-b

since we had \(7-4√3\)in denominator we I multiplied the conjugate \(7+4√3\) in numerator and denominator both so that in denominator I can have the form of \((a+b)*(a-b)=a^2-b^2\)

I hope this helps!!!
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Re: 1/(2-√3)^2=?  [#permalink]

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New post 23 Sep 2018, 15:21
Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

\(\frac{1}{(2 - √3)²}=\)

\(A. 2 + √3\)
\(B. 2 - √3\)
\(C. 7 + 4√3\)
\(D. 7 - 4√3\)
\(E. 4 + 7√3\)


If you aren't sure how to tackle this question, as reasonably fast approach is to try some approximation

By test day, all students should have the following approximations memorized:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2

So, 1/(2 - √3)² ≈ 1/(2 - 1.7
≈ 1/(0.3)²
≈ 1/0.09
11

So, we're looking for an answer choice that equals APPROXIMATELY 11

A. 2 + √3 ≈ 2 + 1.7 = 3.7
B. 2 - √3 ≈ 2 - 1.7 = 0.3
C. 7 + 4√3 ≈ 7 + (4)(1.7) = 13.8
D. 7 - 4√3 ≈ 7 - (4)(1.7) = 0.2
E. 4 + 7√3 ≈ 4 + (7)(1.7) = 15.9

Among the 5 options, answer choice C is best.

Answer: C

Cheers,
Brent
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1/(2-√3)^2=?  [#permalink]

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New post 24 Sep 2018, 06:16
=>

\(\frac{1}{(2-\sqrt{3})^2}= [\frac{1}{(2-\sqrt{3})}]^2= {\frac{(2+\sqrt{3[}{square_root])/{(2-[square_root]3})(2+\sqrt{3})}}^2={\frac{(2+\sqrt{3[}{square_root])/(4-3)}}^2=(2+[square_root]3})^2= 4+4\sqrt{3}+3 = 7+4\sqrt{3}\).

Therefore, the answer is C.
Answer: C
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1/(2-√3)^2=?   [#permalink] 24 Sep 2018, 06:16
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