[24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:

Question

\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}  lies between:

(A) 4 and 5 
(B) 5 and 6 
(C) 6 and 7 
(D) 7 and 8
(E) 8 and 9

Bunuel · Accepted Answer

Square the given expression applying (a+b)^2=a^2+2ab+b^2 : (\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}})^2=(\sqrt{24+5\sqrt{23}})^2+2\sqrt{24+5\sqrt{23}}*\sqrt{24-5\sqrt{23}}+(\sqrt{24-5\sqrt{23}})^2= =24+5\sqrt{23}+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}+24-5\sqrt{23}=48+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})} . Apply (a+b)(a-b)=a^2-b^2 to the second term: 48+2\sqrt{24^2-25*23}=48+2\sqrt{1}=50 . So, our expression equals to: \sqrt{50}\approx{7.something} Answer: D. Similar questions to practice: https://gmatclub.com/forum/square-root-126567.html https://gmatclub.com/forum/root-126569.html https://gmatclub.com/forum/sqrt-15-4-sqr ... 46872.html https://gmatclub.com/forum/topic-101735.html https://gmatclub.com/forum/root-of-138296.html https://gmatclub.com/forum/root-9-root-8 ... 95570.html https://gmatclub.com/forum/if-2-sqrt-5-x ... 05374.html https://gmatclub.com/forum/if-2-root-5-x ... 99663.html https://gmatclub.com/forum/which-of-the- ... ml#p759352 Hope it helps.

mneeti · Answer

Yup, thanks! It helps.

MzJavert · Answer

I took a different approach. The  \sqrt{23}  is between 4 and 5. As 23 is closer to 25 than 16, the square root will be just under 5. So  5\sqrt{23}  will be just under 25. That makes the first part of the equation approximately  \sqrt{49}  and the second part of the equation  \sqrt{1} .

mau5 · Answer

Actually this part is not correct. You can see that the question itself has this term :  \sqrt{24-5\sqrt{23}}  and the term under the root has to be positive. Thus,  24\geq5\sqrt{23} . So, it is just less than 24. With this you can approximate the first part as \approx \sqrt{48}  and not as  \sqrt{49} .

IMO, now you wouldn't know whether the second part is approximately 0 or a little more than 0. So I guess, with this method, one can get stuck between C and D.

I would follow what Bunuel has suggested above.

jlgdr · Answer

First expression is almost 7 since square root of 23 is pretty close to 5 and square root of 49 = 7 By the same token expression two should be very close to 1 since it can't be negative So we have 'Almost 7' + 1 = 'Almost 8' D works just fine Cheers J

vivekisthere · Answer

I am not able to understand how (24^2 - 25*23)^1/2 = 2.

As How I am doing it:

(24^2 - (24+1)(24-1))^1/2
(24^2 - (24^2 - 1^2))^1/2
(24^2 - 24^2)^1/2 by approximation as 24^2-1 would be almost equal to 24^2
=0

Can anyone please help me where I am going wrong?

Many Thanks!!!

KarishmaB · Answer

How do you get that expression two is very close to 1?

\sqrt{24-5\sqrt{23}} 
If we assume that 23 is close to 25 so its square root is 5, (24 - 25) gives you -1. But that is not possible since the whole thing is under a root. It must be slightly positive. Since  \sqrt{23}  is less than 5,  5*\sqrt{23}  must be less than 25, in fact it must be slightly less than 24 since the root has to be 0 or +ve.

So  \sqrt{24-5\sqrt{23}}  will become slightly more than 0.

This means you have 'Almost 7' to which you are adding 'Almost 0'. Can you say whether the sum will be less than or more than 7? No. Hence you get stuck between (C) and (D).

Now to get to the answer you will have to use Bunuel's method.

KarishmaB · Answer

You are misreading something.

\sqrt{24^2 - 25*23} = \sqrt{576 - 575} = 1

The 2 with it is the 2 from the 2ab term of  (a+b)^2 = a^2 + b^2 + 2ab

Put a =  24 + 5\sqrt{23} 
Put b =  24 - 5\sqrt{23}

Now find  (a + b)^2 . You get this as 50. 
Then what is (a+b)? It is  \sqrt{50}

murilomoraes · Answer

For me the best thing to do is note that 5\sqrt{23} is equal to 25*23 doing this you note that the answer to your question will be ˜\sqrt{50} and is between 7 and 8

PareshGmat · Answer

\sqrt{23}  >>>> considered it as  \sqrt{25} 
First expression would come up as 7
So 7 + something would be the Answer = D

mridupawandas · Answer

If we USE approximation then first value we get near 7 and the second value just near 1 so the final value between 7 and 8.

sidoknowia · Answer

Hi,

I solved the question using same method, however it was difficult to judge and so went with C. It might have taken extra time in exam. What to do in similar situations? Any other way around?

LevanKhukhunashvili · Answer

Bunuel  Hi. I've read your explanations and seems like I understood it but can you please tell me what was wrong with my approach?

x= (\sqrt{24+5( 23} ))^(\sqrt{1/2)+(24-5( 23})) )^(1/2) 
1: square both sides -->  X^2 = 24+5(\sqrt{23})+24-5(\sqrt{23}) 
2: positive and negative  5(\sqrt{23}) =0 so  x^2 =24+24=48
3: make a square root from both sides --> x= (\sqrt{48})  and it is less than 7

MrJglass · Answer

Bunuel, I understood your explanation until this point:

Apply  (a+b)(a-b)=a^2-b^2  to the second term:

24+5\sqrt{23}+2\sqrt{24^2-25*23}+24-5\sqrt{23}=48+2=50 .

Is like you decided not to show some of your solving, but I got lost. I don't know how you arrived at the answer. 
Please help, thanks!

Bunuel · Answer

Added extra steps:

(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}})^2=(\sqrt{24+5\sqrt{23}})^2+2\sqrt{24+5\sqrt{23}}*\sqrt{24-5\sqrt{23}}+(\sqrt{24-5\sqrt{23}})^2= 
 =24+5\sqrt{23}+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}+24-5\sqrt{23}=48+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})} .

Apply  (a+b)(a-b)=a^2-b^2  to the second term ( 2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})} ):

48+2\sqrt{24^2-25*23}=48+2\sqrt{1}=50 .

So, our expression equals to:  \sqrt{50}\approx{7.something}

MrJglass · Answer

Thanks a lot!

atova01 · Answer

Hello - one question. I understand how you got to 48+2 = 50 but why did you take the square root of 50? I thought we had gotten rid of the square root of the 24 early on

Bunuel · Answer

First we squared the expression to eliminate the square roots and simplify the expression. Conversely, at the end, we took the square root to balance the initial squaring.

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[24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:

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[24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:

Prep Toolkit

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