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# Root(80)+root(125)

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Math Expert
Joined: 02 Sep 2009
Posts: 47006

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01 Oct 2012, 05:02
2
5
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Difficulty:

5% (low)

Question Stats:

92% (00:38) correct 8% (01:14) wrong based on 1872 sessions

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$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

Practice Questions
Question: 52
Page: 159
Difficulty: 600

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01 Oct 2012, 05:03
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SOLUTION

$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

$$\sqrt{80}+\sqrt{125} =\sqrt{16*5}+\sqrt{25*5}=4\sqrt{5}+5\sqrt{5}=9\sqrt{5}$$.

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01 Oct 2012, 07:43
2
Whe we deal with this problems may be the best thing to do is to break the number into factors

$$\sqrt{80}$$$$=$$$$\sqrt{2*5*2*2*2}$$ ---> $$\sqrt{4*4*5}$$ ---> $$4$$$$\sqrt{5}$$

$$\sqrt{125}$$ $$=$$ $$\sqrt{25 *5}$$ -----> $$5$$ $$\sqrt{5}$$

$$4$$$$\sqrt{5}$$ $$+ 5$$ $$\sqrt{5}$$ $$=$$ $$9$$$$\sqrt{5}$$

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04 Oct 2012, 00:50
2
$$\sqrt{80} = \sqrt{16*5} = 4\sqrt{5}$$ ........(1)

$$\sqrt{125} = \sqrt{25*5} = 5\sqrt{5}$$ .......(2)

Thus $$4\sqrt{5} + 5\sqrt{5} = 9\sqrt{5}$$

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Joined: 12 Jan 2013
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18 Dec 2013, 14:04
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

Practice Questions
Question: 52
Page: 159
Difficulty: 600

The only thing I know is that the answer is just under 24.. And it is definitely not D since D falls between 14 and 15.. And it's clearly not E. And out of the 3 left, B and C are waay too high above 24, and I figured sqrrt of 5 is around "2.something" so A made most sense.

Clearly this approach is very shaky but this time it worked.
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Joined: 13 Feb 2011
Posts: 94

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20 Aug 2014, 23:32
$$\sqrt{80}$$ is approx. 9 and $$\sqrt{125}$$ is approx. 11 so the answer should be close to 20. Choice A is the closest.
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05 Mar 2015, 07:51
I see this question is from a subcategory called "Practice Questions", where can I find that sub?
Math Expert
Joined: 02 Sep 2009
Posts: 47006

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05 Mar 2015, 07:58
erikvm wrote:
I see this question is from a subcategory called "Practice Questions", where can I find that sub?

Those are practice questions from OG13. You can find all questions HERE.
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04 May 2016, 10:03
1
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

Solution:

To solve the problem we must simplify the radicals. Radicals should be simplified whenever possible. Since the square root of a perfect square produces integers, it will often be helpful to locate and simplify perfect squares within a radical expression. Thus, we first locate the perfect squares that divide evenly into 80 and 125, making the simplification of each radical straightforward.

√80 = √16 x √5 = 4√5

√125 = √25 x √5 = 5√5

Now we can add these two results together. Remember to keep the value inside the radical constant and add together the values on the outside.

4√5 + 5√5 = 9√5

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04 May 2016, 10:06
A
80=4*4*5
125=5*5*5
Root(80)=4Root(5)
Similarly 125
hence 9Root(5)
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04 May 2016, 10:19
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

Practice Questions
Question: 52
Page: 159
Difficulty: 600

$$\sqrt{80}+\sqrt{125} =$$ $$4\sqrt{5}+5\sqrt{5}$$ = $$9\sqrt{5}$$

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22 Mar 2017, 23:29
Prime factorization can really help you and so can converting the roots to fraction exponents.

$$\sqrt { 80 } +\sqrt { 125 } =\\ { ({ 2 }^{ 4 } }\times { 5 }^{ 1 })^{ \frac { 1 }{ 2 } }+{ ({ 5 }^{ 2 }\times { 5 }^{ 1 }) }^{ \frac { 1 }{ 2 } }=\\ ({ 2 }^{ \frac { 4 }{ 2 } }\times { 5 }^{ \frac { 1 }{ 2 } })+({ 5 }^{ \frac { 2 }{ 2 } }\times { 5 }^{ \frac { 1 }{ 2 } })\quad =\\ { 2 }^{ 2 }\sqrt { 5 } +5\sqrt { 5 } =\\ 4\sqrt { 5 } +5\sqrt { 5 } =\\ \Rightarrow 9\sqrt { 5 } \\$$
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14 Sep 2017, 22:49
√80 ~~ 9
√ 125 ~~ 11
9+11 = 20
√5 ~~2.1
9*2.1~~20
all other answer not near to this
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Re: Root(80)+root(125)   [#permalink] 14 Sep 2017, 22:49
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