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How many roots does this equation have? \(sqrt(x^2+1) + sqrt(x^2+2) = 2\) (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 Source: GMAT Club Tests  hardest GMAT questions



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12 Aug 2008, 09:43
durgesh79 wrote: How many roots does this equation have?
\(sqrt(x^2+1) + sqrt(x^2+2) = 2\)
0 1 2 3 4 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.
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Re: M12 Q17 [#permalink]
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25 Feb 2010, 12:40
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: \(x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4\) > \(2\sqrt{(x^2+1)(x^2 +2)}=12x^2\). At this point you should square again > \(4x^4+12x^2+8=14x^2+4x^4\) > \(16x^2=7\), 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.
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Re: M12 Q17 [#permalink]
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09 Aug 2008, 09:09
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Square original equation x^2+1 + x^2+2 + 2sqrt[(x^2+1)(x^2+2)] = 4 => 2sqrt[(x^2+1)(x^2+2)] = 1  2x^2
Square again 4(x^4+3x^2+2) = 1 + 4x^4  4x^2 =>8x^2 = 7 => x^2 = 7/8
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.



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abhijit_sen wrote: Square original equation x^2+1 + x^2+2 + 2sqrt[(x^2+1)(x^2+2)] = 4 => 2sqrt[(x^2+1)(x^2+2)] = 1  2x^2
Square again 4(x^4+3x^2+2) = 1 + 4x^4  4x^2 =>8x^2 = 7 => x^2 = 7/8
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"



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14 Aug 2008, 13:09
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IanStewart wrote: durgesh79 wrote: How many roots does this equation have?
\(sqrt(x^2+1) + sqrt(x^2+2) = 2\)
0 1 2 3 4 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.
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TheSituation wrote: Hope it's clear Abudantly clear...thanks. +1 Your posts are incredibly helpful, thank you for taking the time. Just curious, have you written the GMAT yet? If so, how did you score? If you say anything less than 790 I won't believe you.. Thanks again!!!
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Re: M12 Q17 [#permalink]
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25 Feb 2010, 15:05
rbriscoe wrote: 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: \(x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4\) > \(2\sqrt{(x^2+1)(x^2 +2)}=12x^2\). At this point you should square again > \(4x^4+12x^2+8=14x^2+4x^4\) > \(16x^2=7\), 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 > \(2\sqrt{(x^2+1)(x^2 +2)}=12x^2\). At this point you should square again > \((2\sqrt{(x^2+1)(x^2 +2)})^2=(12x^2)^2\) > \(4(x^2+1)(x^2 +2)=14x^2+4x^4\) > \(4x^4+12x^2+8=14x^2+4x^4\), 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. Hope it's clear.
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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



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Re: M12 Q17 [#permalink]
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u don't need to calculate anything. clearly, x^2 >= 0. sqrt 2 = 1.414 so, SQRT(x^2+1) + SQRT(x^2+2) = 2, 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.
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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: \(x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4\) > \(2\sqrt{(x^2+1)(x^2 +2)}=12x^2\). At this point you should square again > \(4x^4+12x^2+8=14x^2+4x^4\) > \(16x^2=7\), 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. this is the kind of questions i had got in real GMAT and i was stumped...took 5 mins, yet no correct answer.. .now i am slowly understanding how to answer these kinds
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Re: M12 Q17 [#permalink]
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06 Mar 2013, 22:48
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Given equation is
\(\sqrt{(x^2 +1)} + \sqrt{(x^2 +2)} = 2\)
As we know that \(x^2\) is greater than zero for all values of \(x\)
So the least possible value of LHS of equation will be when\(x = 0\)
At \(x = 0\), LHS = \(\sqrt{(0 +1)} + \sqrt{(0 +2)} >\) RHS \((2)\)
So equation does not have any real root.....



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Re: M12 Q17 [#permalink]
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21 Aug 2008, 02:42
Thanks guys. We'll change the wording of the question. +1 for everyone.
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Re: M12 Q17 [#permalink]
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25 Feb 2010, 07:39
Quote: Square original equation x^2+1 + x^2+2 + 2sqrt[(x^2+1)(x^2+2)] = 4 => 2sqrt[(x^2+1)(x^2+2)] = 1  2x^2
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.
Answer A. But even answer is right, A



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Re: M12 Q17 [#permalink]
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25 Feb 2010, 10:42
IanStewart wrote: 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.



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Re: M12 Q17 [#permalink]
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25 Feb 2010, 11:58
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?
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Re: M12 Q17 [#permalink]
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25 Feb 2010, 14:12
Hrish88, Pardon my ignorance but what is RHS and LHS? hrish88 wrote: Bunuel wrote: \(x^2 \geq 0\), 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.
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25 Feb 2010, 14:18
rbriscoe wrote: Hrish88,
Pardon my ignorance but what is RHS and LHS? Ha, finally a question I can help out on!! RHS and LHS are referring to righthand side and lefthand side of the equation.
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Re: M12 Q17 [#permalink]
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25 Feb 2010, 14:21
RHS means Right Hand Side and LHS means Left Hand Side of the given equation.
here RHS is 2 and LHS is \(\sqrt{{x^2+1}} + \sqrt{{x^2+2}}\)







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