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98^(1/2) + 72^(1/2) =

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98^(1/2) + 72^(1/2) = [#permalink]

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New post 13 May 2017, 14:53
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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 13 May 2017, 20:06
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Bunuel wrote:
\(\sqrt{98}+ \sqrt{72} =\)

A. \(\sqrt{170}\)

B. \(\sqrt{232}\)

C. \(\sqrt{286}\)

D. \(\sqrt{338}\)

E. \(\sqrt{420}\)


First make the prime factorization:

\(98 = 2 \times 49 = 2 \times 7^2\)
\(72=2 \times 36 =2 \times 6^2 \)

\(\implies \sqrt{98}+\sqrt{72}=\sqrt{2 \times 7^2} + \sqrt{2 \times 6^2 } = 7\sqrt{2} + 6 \sqrt{2}=13\sqrt{2}=\sqrt{13^2 \times 2} = \sqrt{169 \times 2}=\sqrt{338}\)

The answer is D
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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 15 May 2017, 09:24
98−−√+72−−√=98+72=

A. 170−−−√170

B. 232−−−√232

C. 286−−−√286

D. 338−−−√338

E. 420−−−√420
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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 12 Jun 2017, 22:20
Prime factorization of 98 gives 2*(7^2)
Prime factorization of 72 gives (2^3)*(3^2)

From here we can take \(\sqrt{2}\) as a common factor and get D as the answer

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98^(1/2) + 72^(1/2) = [#permalink]

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New post 13 Jun 2017, 04:49
Bunuel wrote:
\(\sqrt{98}+ \sqrt{72} =\)

A. \(\sqrt{170}\)

B. \(\sqrt{232}\)

C. \(\sqrt{286}\)

D. \(\sqrt{338}\)

E. \(\sqrt{420}\)


EASY WAY OUT FOR SUCH QUESTIONS IS APPROXIMATION

\(\sqrt{98}+ \sqrt{72} =9.9+8.6 =\) 18.5

A. \(\sqrt{170} = 13.1 approx.\)

B. \(\sqrt{232}= 15.2 approx.\)

C. \(\sqrt{286}= 16.5 approx.\)

D. \(\sqrt{338}= 18.2 approx.\) CORRECT ANSWER

E. \(\sqrt{420}= 20.5 approx.\)
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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 19 Jun 2017, 22:16
Bunuel wrote:
\(\sqrt{98}+ \sqrt{72} =\)

A. \(\sqrt{170}\)

B. \(\sqrt{232}\)

C. \(\sqrt{286}\)

D. \(\sqrt{338}\)

E. \(\sqrt{420}\)


\sqrt{98} reduces to

49 x 2=
7\sqrt{2} +...

\sqrt{72}
Reduces to 36 x 2
6\sqrt{2}

7\sqrt{2} + 6\sqrt{2} = 13 \sqrt{2}

169 x 2= 338

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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 19 Jun 2017, 22:40
Bunuel wrote:
\(\sqrt{98}+ \sqrt{72} =\)

A. \(\sqrt{170}\)

B. \(\sqrt{232}\)

C. \(\sqrt{286}\)

D. \(\sqrt{338}\)

E. \(\sqrt{420}\)


\(\sqrt{98}+ \sqrt{72}\)

Or, \(7\sqrt{2}+ 6\sqrt{2}\)

Or, \(13\sqrt{2}\)

\(\sqrt{338}\) (When we take 13 inside the square root ), hence answer will be (D)
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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 19 Jun 2017, 23:59
Bunuel wrote:
\(\sqrt{98}+ \sqrt{72} =\)

A. \(\sqrt{170}\)

B. \(\sqrt{232}\)

C. \(\sqrt{286}\)

D. \(\sqrt{338}\)

E. \(\sqrt{420}\)


\(\sqrt{98}+ \sqrt{72}\)
\(= 7\sqrt{2}+ 6\sqrt{2}\)
\(= 13\sqrt{2}\)

\(= \sqrt{338}\)

[Reveal] Spoiler:
Answer D.

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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 20 Jun 2017, 19:12
Bunuel wrote:
\(\sqrt{98}+ \sqrt{72} =\)

A. \(\sqrt{170}\)

B. \(\sqrt{232}\)

C. \(\sqrt{286}\)

D. \(\sqrt{338}\)

E. \(\sqrt{420}\)



Let’s simplify the given expression:

√98 + √72 = √49 x √2 + √36 x √2 = 7√2 + 6√2 = 13√2

Since 13 = √169, 13√2 = √169 x √2 = √338.

Answer: D
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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 20 Jun 2017, 23:32
D
√98+√72=7√2+6√2=13√2=√169*2=√338
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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 19 Jul 2017, 02:43
For these questions I recommend two things
1. Learn Squares upto 30
2. Learn root equivalent
Now Solve
√98+√72
7√2+6√2
√2(7+6)
13√2 by approximation(√2=1.41)
13*1.4=18.2
18=324
19=361
So option D is correct choice
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98^(1/2) + 72^(1/2) = [#permalink]

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New post 01 Oct 2017, 16:10
Bunuel wrote:
\(\sqrt{98}+ \sqrt{72} =\)

A. \(\sqrt{170}\)

B. \(\sqrt{232}\)

C. \(\sqrt{286}\)

D. \(\sqrt{338}\)

E. \(\sqrt{420}\)



easy way

( √98+ √72 ) ^2

=96 + 72 + 2*√(2*49*36*2)

= √338

hope this helps 8-)

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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 04 Oct 2017, 16:45
Bunuel wrote:
\(\sqrt{98}+ \sqrt{72} =\)

A. \(\sqrt{170}\)

B. \(\sqrt{232}\)

C. \(\sqrt{286}\)

D. \(\sqrt{338}\)

E. \(\sqrt{420}\)


√98 + √72 = √49 x √2 + √36 x √2 = 7√2 + 6√2 = 13√2 = √169 x √2 = √338

Answer: D
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Re: 98^(1/2) + 72^(1/2) = [#permalink]

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New post 15 Oct 2017, 12:47
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Top Contributor
Bunuel wrote:
\(\sqrt{98}+ \sqrt{72} =\)

A. \(\sqrt{170}\)

B. \(\sqrt{232}\)

C. \(\sqrt{286}\)

D. \(\sqrt{338}\)

E. \(\sqrt{420}\)


We can also solve this question with some approximation
Here are a few "nice" roots"
√49 = 7
√64 = 8
√81 = 9
√100 = 10

Since √98 is BETWEEN √81 and √100, we know that √98 is BETWEEN 9 and 10
Also, since √98 is VERY CLOSE to √100, we can conclude that √98 is much closer to 10 than it is to 9
So, we might say that √98 ≈ 9.8 or 9.9

Likewise, since √72 is BETWEEN √64 and √81, we know that √72 is BETWEEN 8 and 9
Here, √72 is pretty much halfway between √64 and √81, we might say that √72 ≈ 8.5

NOTE: As we'll soon see, we don't need to be super accurate with our approximations.
We have: √98 + √72 ≈ 9.9 + 8.5 = ≈18.4

Now let's examine a few more "nice" roots"
17² = 289. So, √289 = 17
18² = 324. So, √324 = 18
19² = 361. So, √361 = 19
Since 18.4 is BETWEEN 18 and 19, we're looking for an answer choice that is BETWEEN √324 and √361
Since √338 is the ONLY answer choice BETWEEN √324 and √361, the correct answer must be D


Cheers,
Brent
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Re: 98^(1/2) + 72^(1/2) =   [#permalink] 15 Oct 2017, 12:47
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