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by binit » Wed Jul 22, 2015 8:34 am
Q. If X = sqrt(24 + 5*sqrt(23)) + sqrt(24 - 5*sqrt(23)) then X lies between:
A. 4 & 5
B. 5 & 6
C. 6 & 7
D. 7 & 8
E. 8 & 9


OA after some replies.
Pls explain the best (fastest) way to solve this.

~Binit.
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Source: — Problem Solving |

by DavidG@VeritasPrep » Wed Jul 22, 2015 10:24 am
We're told that x = √(24 + 5√23) + √(24 - 5 √23)

If we square both sides, we'll have x^2= [√(24 + 5√23) + √(24 - 5 √23)]^2

(We just need to remember that we're solving for x^2 now, so we'll have to take the square root to find x later.)

Let's say a = [√(24 + 5√23) and b = √(24 - 5 √23)

We know that (a + b)^2 = a^2 + 2ab + b^2

So [√(24 + 5√23) + √(24 - 5 √23)]^2 = 24 + 5√23 + 2 * √[(24 + 5√23)(24 - 5√23)] + 24 - 5√23

Terms in red above will cancel out to give us 24 + 2 * √[(24 + 5√23)(24 - 5√23)] + 24
Or 48 + 2 * √[(24 + 5√23)(24 - 5√23)]

The terms in blue above can be seen as the difference of squares. If (a + b)(a - b) = a^2 - b^2, then
√[(24 + 5√23)(24 - 5√23)] = (24^)2 - (5√23)^2 = 576 - 25*23 = 576 - 575 = 1.

Substitute 1 in place of [(24 + 5√23)(24 - 5√23)] and we have 48 +2*1 = 50.

If x^2 = 50, then x = √50. If √49 = 7, then √50 must be between 7 and 8.

Answer is D.

(I have seen official questions like this, but the numbers tend to be a tad friendlier.)
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by binit » Wed Jul 22, 2015 10:50 am
Hi Dave, thanks for the detailed solution.

I too have seen problems like this where approximation normally works great. But, when I approximated here: sqrt(23)=4.8, which is very close, I found 5*sqrt(23)=24, which makes the 2nd expression ZERO and X<7.

Do u think its a good idea to do all those steps, as u have shown, in the real test? I don't think it's easy for me under time pressure. Is there any other way?

~Binit.
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by DavidG@VeritasPrep » Wed Jul 22, 2015 11:05 am
Do u think its a good idea to do all those steps, as u have shown, in the real test? I don't think it's easy for me under time pressure. Is there any other way?
It's certainly not unreasonable to estimate here. If you can see that the first term is very close to 7, and the second term is very close 0, but still positive, it's perfectly reasonable to assume that you're dealing with a number slightly greater than 7. The issue, of course, is that when the answer choices are close, as they are here, you might have to solve in order to prove it. (If the correct answer is root 50, which is between 7 and 7.1, it's not hard to imagine someone quickly estimating, say, 6.9.) Let me dig up the official version of the question, so we can assess if the time to solve that one would feel more reasonable.
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by DavidG@VeritasPrep » Wed Jul 22, 2015 11:11 am
Here is the GMATPrep version:

[(√(9 + √80) + √(9 - √80)]^2 =

a) 1
b) 9 - 4√5
c) 18 - 4√5
d) 18
e) 20
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by binit » Wed Jul 22, 2015 11:25 am
I learned my lesson, Dave. I think the question is meant to punish the blind approximation habit.
After understanding ur first method, i can recognize the answer would be E. 20.
Thanks so much for all your help.

~Binit.
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by DavidG@VeritasPrep » Thu Jul 23, 2015 8:20 am
learned my lesson, Dave. I think the question is meant to punish the blind approximation habit.
After understanding ur first method, i can recognize the answer would be E. 20.
Thanks so much for all your help.

~Binit.
Bear in mind that the takeaway is a little more nuanced than this. Estimation, for this question, allows you to eliminate A, B, and C very quickly. They're all way too small. Next, you can see that if we call √80 = 9, we get exactly 18. But √80 isn't exactly 9. So if the answer isn't 18, it's got to be 20.
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by nikhilgmat31 » Wed Jul 29, 2015 3:18 am
DavidG@VeritasPrep wrote:Here is the GMATPrep version:

[(√(9 + √80) + √(9 - √80)]^2 =

a) 1
b) 9 - 4√5
c) 18 - 4√5
d) 18
e) 20
answer is 20 for this question.
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by nikhilgmat31 » Wed Jul 29, 2015 3:20 am
DavidG@VeritasPrep wrote:We're told that x = √(24 + 5√23) + √(24 - 5 √23)

If we square both sides, we'll have x^2= [√(24 + 5√23) + √(24 - 5 √23)]^2

(We just need to remember that we're solving for x^2 now, so we'll have to take the square root to find x later.)

Let's say a = [√(24 + 5√23) and b = √(24 - 5 √23)

We know that (a + b)^2 = a^2 + 2ab + b^2

So [√(24 + 5√23) + √(24 - 5 √23)]^2 = 24 + 5√23 + 2 * √[(24 + 5√23)(24 - 5√23)] + 24 - 5√23

Terms in red above will cancel out to give us 24 + 2 * √[(24 + 5√23)(24 - 5√23)] + 24
Or 48 + 2 * √[(24 + 5√23)(24 - 5√23)]

The terms in blue above can be seen as the difference of squares. If (a + b)(a - b) = a^2 - b^2, then
√[(24 + 5√23)(24 - 5√23)] = (24^)2 - (5√23)^2 = 576 - 25*23 = 576 - 575 = 1.

Substitute 1 in place of [(24 + 5√23)(24 - 5√23)] and we have 48 +2*1 = 50.

If x^2 = 50, then x = √50. If √49 = 7, then √50 must be between 7 and 8.

Answer is D.

(I have seen official questions like this, but the numbers tend to be a tad friendlier.)
for easier estimation

we can write √50 as 5√2 & 5 * 1.41 ~ 7.05
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