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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

I don't think we need to do any algebra here: \(x^2 \geq 0\), so \(\sqrt{x^2+1} + \sqrt{x^2+2} \geq 1 + \sqrt{2}\), which is larger than 2. Hence no (real) solutions for x. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

I got A but solved thru a different (less clever) method. I think I may have answered correct in spite of myself though, can someone tell me if my solution is mathmatically sound or if I arrived at the correct solution by luck.

Begin by squaring both sides

(x^2+1) + (x^2 +2) = 4 drop brackets and solve 2x^2=1 x^2=1/2 x = sq root of 1/2 therefore no real roots, therefore A

feedback?

This would be the longer way, plus you'll need to square twice not once, as you made a mistake while squaring first time.

\((a+b)^2=a^2+2ab+b^2\), so when you square both sides you'll get: \(x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4\) --> \(2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2\). At this point you should square again --> \(4x^4+12x^2+8=1-4x^2+4x^4\) --> \(16x^2=-7\), no real \(x\) satisfies this equation.

There is one more problem with your solution:

You've got (though incorrectly): \(x^2=\frac{1}{2}\) and then you concluded that this equation has no real roots, which is not right. This quadratic equation has TWO real roots: \(x=\frac{1}{\sqrt{2}}\) and \(x=-{\frac{1}{\sqrt{2}}}\). Real roots doesn't mean that the roots must be integers, real roots means that roots must not be complex numbers, which I think we shouldn't even mention as GMAT deals ONLY with real numbers. For example \(x^2=-1\) has no real roots and for GMAT it means that this equation has no roots, no need to consider complex roots and imaginary numbers.

As x^2 is -ve, so this equation cannot have solution for any real value of x. Although complex numbers solutions can be obtained, these are not part of regular GMAT.

I don't think we need to do any algebra here: \(x^2 \geq 0\), so \(\sqrt{x^2+1} + \sqrt{x^2+2} \geq 1 + \sqrt{2}\), which is larger than 2. Hence no (real) solutions for x.

sorry but i didnt understand this can anyone explain.

\(x^2 \geq 0\), means that the lowest value of LHS is when \(x=0\) --> \(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt2=2.4>RHS=2\), hence as the lowest possible value of LHS is still greater than RHS, the equation has no real roots.

As x^2 is -ve, so this equation cannot have solution for any real value of x. Although complex numbers solutions can be obtained, these are not part of regular GMAT.

Answer A.

thanks Abhijit, technically the language of the question is worng, should be changed to "How many real roots does this equation have"

I don't think we need to do any algebra here: \(x^2 \geq 0\), so \(\sqrt{x^2+1} + \sqrt{x^2+2} \geq 1 + \sqrt{2}\), which is larger than 2. Hence no (real) solutions for x.

Agreed .. good point. +1 for you.

Durgesh, Question wording need to be changed to real roots to make it clear. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

don't think we need hard algebra, just under the square should be x^2+1>0 or x^2+2>0 so, x^2>-1 incorrect x^2>-2 incorrect thus no real roots... answer is A) 0

it doesn't have any solution beacuse left hand side is always greater than 2. the min value of left hand side is 2.414. which is greater than 2. _________________

kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

Square again 4(x^4+3x^2+2) = 1 + 4x^4 - 4x^2 =>8x^2 = -7 should be 16x^2=-7 => x^2 = -7/8 then x^2=-7/16

As x^2 is -ve, so this equation cannot have solution for any real value of x. Although complex numbers solutions can be obtained, these are not part of regular GMAT.

I don't think we need to do any algebra here: \(x^2 \geq 0\), so \(\sqrt{x^2+1} + \sqrt{x^2+2} \geq 1 + \sqrt{2}\), which is larger than 2. Hence no (real) solutions for x.

sorry but i didnt understand this can anyone explain.

I got A but solved thru a different (less clever) method. I think I may have answered correct in spite of myself though, can someone tell me if my solution is mathmatically sound or if I arrived at the correct solution by luck.

Begin by squaring both sides

(x^2+1) + (x^2 +2) = 4 drop brackets and solve 2x^2=1 x^2=1/2 x = sq root of 1/2 therefore no real roots, therefore A

\(x^2 \geq 0\), means that the lowest value of LHS is when \(x=0\) --> \(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt2=2.4>RHS=2\), hence as the lowest possible value of LHS is still greater than RHS, the equation has no real roots.

Hope it's clear.

Thanks Bunuel.

_________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own. I am the force; I can clear any obstacle before me or I can be lost in the maze. My choice; my responsibility; win or lose, only I hold the key to my destiny - Elaine Maxwell

I am sooo lost on this question. I solved it the same as TheSiuation. I am do not understand your method below. Although I am sure it is pbvious to everyone but please clarify the rule, therum or postulate that you followed to solve the equation. Thanks in advacne.

Bunuel wrote:

TheSituation wrote:

I got A but solved thru a different (less clever) method. I think I may have answered correct in spite of myself though, can someone tell me if my solution is mathmatically sound or if I arrived at the correct solution by luck.

Begin by squaring both sides

(x^2+1) + (x^2 +2) = 4 drop brackets and solve 2x^2=1 x^2=1/2 x = sq root of 1/2 therefore no real roots, therefore A

feedback?

This would be the longer way, plus you'll need to square twice not once, as you made a mistake while squaring first time.

\((a+b)^2=a^2+2ab+b^2\), so when you square both sides you'll get: \(x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4\) --> \(2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2\). At this point you should square again --> \(4x^4+12x^2+8=1-4x^2+4x^4\) --> \(16x^2=-7\), no real \(x\) satisfies this equation.

There is one more problem with your solution:

You've got (though incorrectly): \(x^2=\frac{1}{2}\) and then you concluded that this equation has no real roots, which is not right. This quadratic equation has TWO real roots: \(x=\frac{1}{\sqrt{2}}\) and \(x=-{\frac{1}{\sqrt{2}}}\). Real roots doesn't mean that the roots must be integers, real roots means that roots must not be complex numbers, which I think we shouldn't even mention as GMAT deals ONLY with real numbers. For example \(x^2=-1\) has no real roots and for GMAT it means that this equation has no roots, no need to consider complex roots and imaginary numbers.

Hope it's clear.

_________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own. I am the force; I can clear any obstacle before me or I can be lost in the maze. My choice; my responsibility; win or lose, only I hold the key to my destiny - Elaine Maxwell

I am sooo lost on this question. I solved it the same as TheSiuation. I am do not understand your method below. Although I am sure it is pbvious to everyone but please clarify the rule, therum or postulate that you followed to solve the equation. Thanks in advacne.

Bunuel wrote:

TheSituation wrote:

I got A but solved thru a different (less clever) method. I think I may have answered correct in spite of myself though, can someone tell me if my solution is mathmatically sound or if I arrived at the correct solution by luck.

Begin by squaring both sides

(x^2+1) + (x^2 +2) = 4 drop brackets and solve 2x^2=1 x^2=1/2 x = sq root of 1/2 therefore no real roots, therefore A

feedback?

This would be the longer way, plus you'll need to square twice not once, as you made a mistake while squaring first time.

\((a+b)^2=a^2+2ab+b^2\), so when you square both sides you'll get: \(x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4\) --> \(2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2\). At this point you should square again --> \(4x^4+12x^2+8=1-4x^2+4x^4\) --> \(16x^2=-7\), no real \(x\) satisfies this equation.

There is one more problem with your solution:

You've got (though incorrectly): \(x^2=\frac{1}{2}\) and then you concluded that this equation has no real roots, which is not right. This quadratic equation has TWO real roots: \(x=\frac{1}{\sqrt{2}}\) and \(x=-{\frac{1}{\sqrt{2}}}\). Real roots doesn't mean that the roots must be integers, real roots means that roots must not be complex numbers, which I think we shouldn't even mention as GMAT deals ONLY with real numbers. For example \(x^2=-1\) has no real roots and for GMAT it means that this equation has no roots, no need to consider complex roots and imaginary numbers.

Hope it's clear.

Frankly speaking I solved this question in different manner, as shown in my first post in this thread. The above solution is also valid, however it's quite time consuming.

We have: \(\sqrt{x^2+1}+\sqrt{x^2+2}=2\). Now, if you choose to square both sides, then when squaring LHS \(\sqrt{x^2+1}+\sqrt{x^2+2}\), you should apply the formula: \((a+b)^2=a^2+2ab+b^2\) --> \((\sqrt{x^2+1})^2+2{(\sqrt{x^2+1})}{(\sqrt{x^2+2})}+(\sqrt{x^2+2})^2=x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2\). When squaring RHS yuo'll get \(2^2=4\).

So you'll get: \(x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2=4\), rearrange --> \(2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2\). At this point you should square again --> \((2\sqrt{(x^2+1)(x^2 +2)})^2=(1-2x^2)^2\) --> \(4(x^2+1)(x^2 +2)=1-4x^2+4x^4\) --> \(4x^4+12x^2+8=1-4x^2+4x^4\), rearrange again, \(4x^4\) will cancel out --> \(16x^2=-7\) --> \(x^2=-\frac{7}{16}\). Now, \(x^2\) can not equal to negative number, which means that there doesn't exist an \(x\) to satisfies the given equation. So, no roots.