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M12-17

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M12-17  [#permalink]

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New post 16 Sep 2014, 00:47
1
4
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

64% (01:56) correct 36% (01:36) wrong based on 47 sessions

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Re M12-17  [#permalink]

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New post 16 Sep 2014, 00:47
1
2
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
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Re: M12-17  [#permalink]

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New post 08 Oct 2014, 06:36
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: M12-17  [#permalink]

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New post 08 Oct 2014, 16:24
Awesome approach...I tried the same thing...squared on both sides and then got spiraled within the roots...
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Re: M12-17  [#permalink]

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New post 22 Oct 2014, 19:23
3
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: M12-17  [#permalink]

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New post 03 Apr 2016, 04:58
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: M12-17  [#permalink]

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New post 11 Dec 2016, 07:56
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^2-4ac) 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: M12-17  [#permalink]

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New post 16 Dec 2016, 08:50
1
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: M12-17  [#permalink]

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New post 16 Dec 2016, 12:33
jahidhassan wrote:
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.


By substituting x=0 into the equation.
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Re: M12-17  [#permalink]

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New post 25 May 2017, 15:25
3
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
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M12-17  [#permalink]

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New post 05 Oct 2017, 14:43
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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
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Re: M12-17  [#permalink]

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New post 09 Jun 2018, 11:16
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
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New post 21 Nov 2018, 07:57
Bunuel wrote:
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



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

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

let \({x^2 + 1}\) = y

\(\sqrt{y} + \sqrt{y + 1} = 2\)-------------------------------(1)

squaring both side

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

2y+1 + \(2(\sqrt{y} * \sqrt{y + 1} )\) = 4

2y + \(2(\sqrt{y} * \sqrt{y + 1} )\) = 3

2 (y + \((\sqrt{y} * \sqrt{y + 1} )\) ) = 3

2 \(\sqrt{y}\) ( \(\sqrt{y}+ \sqrt{y + 1}\) ) =3

2 \(\sqrt{y}\) (2) =3-------------------------------------------( from 1)

4 \(\sqrt{y}\) =3

\(\sqrt{y}\) =\(\frac{3}{4}\)

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

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

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

\({x^2 + 1}\) = \(\frac{9}{16}\)

\({x^2 }\) =\(\frac{9}{16} -1\)

\({x^2 }\) = -ve value

so no Solution for X. so ans is A
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Re M12-17  [#permalink]

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New post 20 Feb 2019, 20:09
I think this is a high-quality question and I agree with explanation.
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New post 14 Apr 2019, 03:18
I got this x^2= -ve value

So equation has 2 complex roots ? Bunuel
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New post 14 Apr 2019, 03:25
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Re: M12-17   [#permalink] 14 Apr 2019, 03:25
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