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555-605 (Medium)|   Exponents|   Roots|      
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MHIKER
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{1/(4-√t15)}^2= Updated on: 04 Dec 2020, 10:17
Originally posted by MHIKER on 18 Jan 2012, 09:27.
Last edited by MHIKER on 04 Dec 2020, 10:17, edited 1 time in total.
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C
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25% (medium)

Question Stats:

75% (01:36) correct 25% (01:57) wrong based on 1318 sessions

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\((\frac{1}{{4-√15}})^2 \)=

a. 16+16√15
b. 31- 8√15
c. 31+ 8√15
d. 32- 4√15
e. 32 + 4√15
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C
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{1/(4-√t15)}^2= 18 Jan 2012, 10:40
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{1/(4-√15)}^2 =

a. 16+16√15
b. 31- 8√15
c. 31+ 8√15
d. 32- 4√15
e. 32 + 4√15
Show more

Multiply both nominator and denominator by \(4+\sqrt{15}\) and then apply \((a-b)(a+b)=a^2-b^2\) for denominator: \((\frac{1}{4-\sqrt{15}})^2=(\frac{4+\sqrt{15}}{(4-\sqrt{15})(4+\sqrt{15})})^2=(\frac{4+\sqrt{15}}{16-15})^2=16+8\sqrt{15}+15=31+8\sqrt{15}\).

Answer: C.
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PareshGmat
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{1/(4-√t15)}^2= 27 Feb 2014, 22:03
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Did in this way:

1 / (16 - 8 root15 + 15)

= 1 / (31 - 8 root15)

Multiply numerator & denominator by (31 + 8 root15)

= (31 + 8 root15) / (961-960) = (31 + 8 root15)

Answer = C
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{1/(4-√t15)}^2= Updated on: 30 Dec 2020, 14:50
Originally posted by MHIKER on 04 Dec 2020, 10:29.
Last edited by MHIKER on 30 Dec 2020, 14:50, edited 4 times in total.
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MHIKER
\((\frac{1}{{4-√15}})^2 \)=

a. 16+16√15
b. 31- 8√15
c. 31+ 8√15
d. 32- 4√15
e. 32 + 4√15
Show more


Multiplying, the numerator and denominator by \((4+\sqrt{15})\), applying \(a^2-b^2\) and \((a+b)^2\) formulae.

\(=\frac{{16+8\sqrt{15}+15}}{16-15}\)

\(31+8\sqrt{15}\)

The answer is \(C\)
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{1/(4-√t15)}^2= 04 Dec 2020, 11:32
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MHIKER
\((\frac{1}{{4-√15}})^2 \)=

a. 16+16√15
b. 31- 8√15
c. 31+ 8√15
d. 32- 4√15
e. 32 + 4√15
Show more
\((\frac{1}{{4-√15}})^2 \)

\(=(\frac{(4+√15)}{(4-√15)(4+√15)})^2 \)

\(=(\frac{16+8√15+15}{16-15}) \)

\(= 31+8√15\), Answer is (C)
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{1/(4-√t15)}^2= 05 May 2022, 06:42
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MHIKER
\((\frac{1}{4-\sqrt{15}})^2\)=

a. 16+16√15
b. 31- 8√15
c. 31+ 8√15
d. 32- 4√15
e. 32 + 4√15
Show more

Strategy: For this question, we can either expand \((\frac{1}{4-\sqrt{15}})^2\) and then "fix" the resulting denominator, or " fix" the denominator first and then expand.
Let's fix the denominator first

Take: \(\frac{1}{4-\sqrt{15}}\)

Multiply numerator and denominator by \(4+\sqrt{15}\), which is the complement of \(4-\sqrt{15}\)

When we do this we get: \(\frac{1}{4-\sqrt{15}} = \frac{(1)(4+\sqrt{15})}{(4-\sqrt{15})(4+\sqrt{15})}= \frac{4+\sqrt{15}}{16 - 15}= \frac{4+\sqrt{15}}{1} = 4+\sqrt{15}\)

Substitute this "fixed" expression into the original expression to get: \((\frac{1}{4-\sqrt{15}})^2 = (4+\sqrt{15})^2 = (4+\sqrt{15})(4+\sqrt{15}) = 16 + 4\sqrt{15} + 4\sqrt{15} + (\sqrt{15})^2 = 31 + 8\sqrt{15}\)

Answer: C
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{1/(4-√t15)}^2= 23 Sep 2023, 10:47
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\(\left[ \begin{array}{cc|r} \dfrac{1}{4-\sqrt{15}} \end{array} \right]^2\)

This problem is based on concept of Roots called as Conjugate Roots.

In these kind of problem to simplify the expression (and the denominator) we multiply the numerator and the denominator with the Conjugate of the denominator

Conjugate of \(4-\sqrt{15}\) is \(4+\sqrt{15}\) (Same expression with sign of the square root term reversed)

=> \((\frac{1}{4-\sqrt{15}})^2\) = \((\frac{1}{4-\sqrt{15}} * \frac{4+\sqrt{15}}{4+\sqrt{15}})^2\)
= \((\frac{4+\sqrt{15}}{(4-\sqrt{15})*(4+\sqrt{15})})^2\)

Using (a-b)*(a+b) = \(a^2 - b^2\) in the denominator where a = 4 and b = \(\sqrt{15}\), we get

= \((\frac{4+\sqrt{15}}{(4^2)-15})^2\)
= \((\frac{4+\sqrt{15}}{16-15})^2\)
= \((\frac{4+\sqrt{15}}{1})^2\)
= \((4+\sqrt{15})^2\)
= \(4^2 + 2*4*\sqrt{15} + 15\)
= 31 + 8\(\sqrt{15}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Rationalize Roots

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{1/(4-√t15)}^2= 12 Apr 2026, 02:50
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