Last visit was: 14 Jul 2025, 14:29 It is currently 14 Jul 2025, 14:29
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,570
Own Kudos:
Given Kudos: 98,182
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,570
Kudos: 741,403
 [38]
5
Kudos
Add Kudos
33
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,570
Own Kudos:
Given Kudos: 98,182
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,570
Kudos: 741,403
 [18]
9
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
avatar
illthinker
Joined: 17 Sep 2017
Last visit: 28 Aug 2019
Posts: 6
Own Kudos:
36
 [13]
Given Kudos: 5
Concentration: Finance, Entrepreneurship
GMAT 1: 730 Q48 V42
GPA: 3.8
GMAT 1: 730 Q48 V42
Posts: 6
Kudos: 36
 [13]
11
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
avatar
p2bhokie
Joined: 23 Jan 2012
Last visit: 27 Sep 2015
Posts: 38
Own Kudos:
66
 [1]
Given Kudos: 40
Posts: 38
Kudos: 66
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Awesome approach...I tried the same thing...squared on both sides and then got spiraled within the roots...
avatar
spence11
Joined: 06 Apr 2017
Last visit: 11 Apr 2018
Posts: 20
Own Kudos:
122
 [21]
Given Kudos: 38
Location: United States (OR)
Concentration: Finance, Leadership
GMAT 1: 730 Q49 V40
GPA: 3.98
WE:Corporate Finance (Healthcare/Pharmaceuticals)
GMAT 1: 730 Q49 V40
Posts: 20
Kudos: 122
 [21]
15
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
The Official Solution already posted is the kind of logic that is given a handy reward on the GMAT, but for those of you who started to simplify, I wanted to show an alternative route to getting this correct.

\(\sqrt{x^2+1}+\sqrt{x^2+2} = 2\)

\((\sqrt{x^2+1}+\sqrt{x^2+2})^2 = 2^2\)

\(x^2+1+2(\sqrt{x^2+1})(\sqrt{x^2+2})+x^2+2=4\)

\(2x^2+2(\sqrt{x^4+3x^2+2})=1\)

\(2(x^2+\sqrt{x^4+3x^2+2})=1\)

\(x^2+\sqrt{x^4+3x^2+2}=\frac{1}{2}\)

\(\sqrt{x^4+3x^2+2}=\frac{1}{2}-x^2\)

\(x^4+3x^2+2=(\frac{1}{2}-x^2)^2\)

\(x^4+3x^2+2=\frac{1}{4}-x^2+x^4\)

\(4x^2+\frac{7}{4}=0 => 16x^2+7=0\)

This is a quadratic equation of the form \(ax^2+bx+c=0\), where \(a=16\), \(b=0\), and \(c=7\)

The discriminant of a quadratic equation is given by the equation \(D=b^2-4ac\)

When \(D>0\), there are exactly two real solutions
When \(D=0\), there is exactly one real solutions
When \(D<0\), there are no real solutions

In this case \(D=0^2-4(16)(7)<0\), so there are no real solutions.

Answer A
User avatar
lukachiri
Joined: 17 Apr 2018
Last visit: 16 Dec 2021
Posts: 23
Own Kudos:
6
 [1]
Given Kudos: 41
Location: India
GMAT 1: 710 Q49 V38
GPA: 4
WE:Engineering (Energy)
GMAT 1: 710 Q49 V38
Posts: 23
Kudos: 6
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hey Bunuel !
This Equation will have two complex roots +i(√7/4) and -i(√7/4).
Since it is not mentioned in the problem that we have to consider only real roots, i marked C as the answer.
Am i missing something ? :dazed
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,570
Own Kudos:
741,403
 [3]
Given Kudos: 98,182
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,570
Kudos: 741,403
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
fatehghadeer
hey Bunuel !
This Equation will have two complex roots +i(√7/4) and -i(√7/4).
Since it is not mentioned in the problem that we have to consider only real roots, i marked C as the answer.
Am i missing something ? :dazed

Numbers on the GMAT are real by default (GMAT deals with only real numbers), so no need to consider complex roots.
User avatar
Raxit85
Joined: 22 Feb 2018
Last visit: 13 Jul 2025
Posts: 768
Own Kudos:
Given Kudos: 135
Posts: 768
Kudos: 1,140
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Official Solution:


How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?


A. 0
B. 1
C. 2
D. 3
E. 4


We know that \(x^2 \ge 0\) so the least value of the left hand side (LHS) of the equation is for \(x=0\):

\(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\). Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real \(x\) can satisfy given equation.


Answer: A

Bunuel VeritasKarishma,

Can you please let me know the concept involved in your approach?

In case of LHS = RHS or LHS<RHS, then what will be inferred?

Regards!!
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 14 Jul 2025
Posts: 16,106
Own Kudos:
74,320
 [7]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,106
Kudos: 74,320
 [7]
7
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Raxit85
Bunuel
Official Solution:


How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?


A. 0
B. 1
C. 2
D. 3
E. 4


We know that \(x^2 \ge 0\) so the least value of the left hand side (LHS) of the equation is for \(x=0\):

\(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\). Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real \(x\) can satisfy given equation.


Answer: A

Bunuel VeritasKarishma,

Can you please let me know the concept involved in your approach?

In case of LHS = RHS or LHS<RHS, then what will be inferred?

Regards!!

'Roots of the equation' means the values of x for which the equation will be true.

\(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\)

The point here is that the equation CANNOT be true for any values of x.

When I look at the equation, I don't feel like squaring because even after squaring, I will need to square again to get rid of the square root. So instead I notice the 2 on the RHS and wonder what values the square roots on LHS can take. Since square roots will be positive only, I wonder whether each can be 1. Then I notice that inside the square root, I have x^2 which must be 0 or positive and 1. So the first square root can be 1. I notice the second square root and see that x^2 must be 0 or positive and there is a 2. So the second square root must be at least 1.4 (sqrt2).
Since everything is all positive, I see that LHS must be at least 2.4 (when x = 0). Then can LHS be 2? No. It cannot be for ANY value of x. So for NO value of x will the equation be satisfied.
Hence number of roots = 0.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,570
Own Kudos:
Given Kudos: 98,182
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,570
Kudos: 741,403
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
User avatar
r_putra_rp
Joined: 03 May 2023
Last visit: 16 Jul 2024
Posts: 22
Own Kudos:
5
 [1]
Location: Indonesia
Schools: Sloan '26
GRE 1: Q166 V163
GPA: 3.01
Schools: Sloan '26
GRE 1: Q166 V163
Posts: 22
Kudos: 5
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Amazing and beautiful question indeed!

The equation is indeed jarring and scary, and is tempting to try to solve by squaring or taking roots (I know I did, tried for almost 3 minutes :dazed :dazed :cry: )

But after reading again, I find a 30 seconds solution by realizing that this is not an algebra but a number property/theory problem

reiterating the question
\(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\)

Let's think for a sec. Okay, so we need two roots of a number that equals 2. Fair enough.

Also notice that both of the terms have an \(x^2\) term, which is always positive or zero, \(x>=0\). Up to this point, they're not helping at all.

But wait! Each of the terms are always greater than 1!
For the first term \(\sqrt{x^2 + 1} \), min value is 1 when \(x^2 = 0\)
For the second term \(\sqrt{x^2 + 2} \), min value is \(\sqrt{2} \approx 1.4 \) when \(x^2 = 0\).

You dont actually need the approximate number of 1.4 to solve this problem, only that obviously \(\sqrt{2} > \sqrt{1} \) or that \(\sqrt{2}\) MUST BE greater than 1

Now, that since the first square root term is 1 and the second is greater than 1, adding both CAN NOT BE EQUAL to 2, thus it doesn't have any solution in real numbers, hence A
User avatar
BottomJee
User avatar
Retired Moderator
Joined: 05 May 2019
Last visit: 09 Jun 2025
Posts: 996
Own Kudos:
Given Kudos: 1,009
Affiliations: GMAT Club
Location: India
GMAT Focus 1: 645 Q82 V81 DI82
GMAT 1: 430 Q31 V19
GMAT 2: 570 Q44 V25
GMAT 3: 660 Q48 V33
GPA: 3.26
WE:Engineering (Manufacturing)
Products:
GMAT Focus 1: 645 Q82 V81 DI82
GMAT 3: 660 Q48 V33
Posts: 996
Kudos: 1,213
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I think this is a high-quality question and I agree with explanation.
User avatar
sujoykrdatta
Joined: 26 Jun 2014
Last visit: 13 Jul 2025
Posts: 539
Own Kudos:
Given Kudos: 13
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
Expert
Expert reply
Posts: 539
Kudos: 1,043
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?


A. 0
B. 1
C. 2
D. 3
E. 4


\(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) ... (1)

Observe that:
\(\{sqrt{x^2 + 1}}^2 - \{sqrt{x^2 + 2}}^2 = x^2 +2 - (x^2 - 1) = 1\)

Dividing this by the give equation and using a²-b²=(a+b)(a-b), we have:

\(\sqrt{x^2 + 1} - \sqrt{x^2 + 2} = 1/2\) ... (2)

Add 1 and 2:
2 * sqrt{x^2 + 2} = 2 + 1/2 = 5/2
=> sqrt{x^2 + 2} = 5/4
=> x² + 2 = 25/16
=> x² = 25/16 - 2 < 0
Thus, there is no real solution

Answer A

Posted from my mobile device
User avatar
Manvi01
Joined: 09 Jul 2024
Last visit: 25 May 2025
Posts: 11
Given Kudos: 225
Location: India
Posts: 11
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I like the solution - it’s helpful.
User avatar
scrantonstrangler
Joined: 19 Feb 2022
Last visit: 12 Jul 2025
Posts: 117
Own Kudos:
51
 [1]
Given Kudos: 69
Status:Preparing for the GMAT
Location: India
GMAT 1: 700 Q49 V35
GPA: 3.33
WE:Consulting (Consulting)
GMAT 1: 700 Q49 V35
Posts: 117
Kudos: 51
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This is a great question that’s helpful for learning.
User avatar
Premkungg
Joined: 19 Apr 2022
Last visit: 29 Mar 2025
Posts: 13
Given Kudos: 95
Posts: 13
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I like the solution - it’s helpful.
User avatar
avellanjc
Joined: 03 Jan 2025
Last visit: 14 Jul 2025
Posts: 19
Own Kudos:
Given Kudos: 5
Products:
Posts: 19
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi,

The 1st line of this solution shouldnt have sq roots \sqrt{}.

He is letting A=X^2 +1 and B= X^2 +2

Rest is OK.

Regards,

Juan C. Avellan


illthinker
Simpler version of the mere-mortal approach:

Let \(\sqrt{x^2+1} = A\) and let \(\sqrt{x^2+2} = B\) \(=>\)
\(\sqrt{A} + \sqrt{B} = 2\)
Rearrange and square both sides \(=>\)
\(A = (2-\sqrt{B})(2-\sqrt{B})\)
\(A = 4-4\sqrt{B}+B\)
Substitute original expressions back in \(=>\)
\(x^2+1 = 4-4\sqrt{x^2+2}+x^2+2\)
\(4\sqrt{x^2+2}=5\)
Square both sides, rearrange \(=>\)
\(16x^2=-7\) which is not true for any real x, so 0 solutions
Moderators:
Math Expert
102570 posts
Founder
41097 posts