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# M12-17

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Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128871 [0], given: 12183

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16 Sep 2014, 00:47
Expert's post
7
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

63% (01:09) correct 37% (01:28) wrong based on 35 sessions

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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
[Reveal] Spoiler: OA

_________________

Kudos [?]: 128871 [0], given: 12183

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128871 [0], given: 12183

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16 Sep 2014, 00:47
Expert's post
1
This post was
BOOKMARKED
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.

_________________

Kudos [?]: 128871 [0], given: 12183

Current Student
Joined: 20 Jan 2014
Posts: 175

Kudos [?]: 69 [0], given: 120

Location: India
Concentration: Technology, Marketing

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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.

Awkward question. I was trying to square both site and ultimately got stuck somewhere.
Nice approach to this ques.
_________________

Kudos [?]: 69 [0], given: 120

Manager
Joined: 23 Jan 2012
Posts: 75

Kudos [?]: 15 [0], given: 40

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08 Oct 2014, 16:24
Awesome approach...I tried the same thing...squared on both sides and then got spiraled within the roots...

Kudos [?]: 15 [0], given: 40

Intern
Joined: 22 Jun 2014
Posts: 22

Kudos [?]: 12 [2], given: 12

Concentration: General Management, Finance
GMAT 1: 700 Q50 V34
GRE 1: 1400 Q800 V600
GPA: 3.68

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22 Oct 2014, 19:23
2
KUDOS
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

Kudos [?]: 12 [2], given: 12

Manager
Joined: 03 Aug 2015
Posts: 64

Kudos [?]: 8 [0], given: 219

Concentration: Strategy, Technology
Schools: ISB '18, SPJ GMBA '17
GMAT 1: 680 Q48 V35

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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.

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

Kudos [?]: 8 [0], given: 219

Intern
Joined: 27 Oct 2015
Posts: 20

Kudos [?]: [0], given: 8

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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.

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

Kudos [?]: [0], given: 8

Intern
Joined: 23 Jun 2016
Posts: 18

Kudos [?]: [0], given: 36

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16 Dec 2016, 08:50
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.

Hi,
Would you please explain how $$\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4$$ is derived.

Thanks.

Kudos [?]: [0], given: 36

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128871 [0], given: 12183

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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.

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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Kudos [?]: 128871 [0], given: 12183

Intern
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Posts: 27

Kudos [?]: 7 [0], given: 38

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25 May 2017, 15:25
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 $$D=b^2-4ac$$

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.

Kudos [?]: 7 [0], given: 38

Intern
Joined: 17 Sep 2017
Posts: 3

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05 Oct 2017, 14:43
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

Kudos [?]: 0 [0], given: 2

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

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