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Re M1217 [#permalink]
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15 Sep 2014, 23:47



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Re: M1217 [#permalink]
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08 Oct 2014, 05:36
Bunuel wrote: 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 Awkward question. I was trying to square both site and ultimately got stuck somewhere. Nice approach to this ques.
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Re: M1217 [#permalink]
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08 Oct 2014, 15:24
Awesome approach...I tried the same thing...squared on both sides and then got spiraled within the roots...



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Re: M1217 [#permalink]
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22 Oct 2014, 18:23
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easy approach. but for ppl who have been trying to square. What can be done is to take rt(x2+2) to the right side of the equation. Square it and solve for roots. x2 turns out to be negative. Hence no solution



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Re: M1217 [#permalink]
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03 Apr 2016, 03:58
Bunuel wrote: 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 Bunel, Pls could you tell how to approach a Q asking to find no. of roots? I would like to know the theory and approach? Thanks, A



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Re: M1217 [#permalink]
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11 Dec 2016, 06:56
Hi Buenel,
We know that x2≥0x2≥0 so the least value of the left hand side (LHS) of the equation is for x=0x=0:
x2+1−−−−−√+x2+2−−−−−√=1+2√≈2.4x2+1+x2+2=1+2≈2.4. Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real xx can satisfy given equation.
The correct answer is A
My Doubt is that since the question is asking for roots, dont we have to find the discriminant (b^24ac) and then determine the no. of roots?
I tried finding the discriminant as follows:
square both sides
b^2 = 4 a= 2 c = 1
Therefor b^2  4ac = 4+8 = 12.
Please tell me where I am making an error



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Re: M1217 [#permalink]
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16 Dec 2016, 07:50
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Bunuel wrote: 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 Hi, Would you please explain how \(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\) is derived. Thanks.



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Re: M1217 [#permalink]
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16 Dec 2016, 11:33



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Re: M1217 [#permalink]
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25 May 2017, 14:25
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 \(D=b^24ac\)
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 \(D=0^24(16)(7)<0\), so there are no real solutions.
Answer A



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Simpler version of the meremortal approach:
Let \(\sqrt{x^2+1} = A\) and let \(\sqrt{x^2+2} = B\) \(=>\) \(\sqrt{A} + \sqrt{B} = 2\) Rearrange and square both sides \(=>\) \(A = (2\sqrt{B})(2\sqrt{B})\) \(A = 44\sqrt{B}+B\) Substitute original expressions back in \(=>\) \(x^2+1 = 44\sqrt{x^2+2}+x^2+2\) \(4\sqrt{x^2+2}=5\) Square both sides, rearrange \(=>\) \(16x^2=7\) which is not true for any real x, so 0 solutions










