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# M12-17

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Math Expert
Joined: 02 Sep 2009
Posts: 49271

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16 Sep 2014, 00:47
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Difficulty:

75% (hard)

Question Stats:

56% (01:14) correct 44% (01:13) wrong based on 50 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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 49271

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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.

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Manager
Joined: 20 Jan 2014
Posts: 159
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.
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Joined: 23 Jan 2012
Posts: 68

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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...
Intern
Joined: 22 Jun 2014
Posts: 22
Concentration: General Management, Finance
GMAT 1: 700 Q50 V34
GRE 1: Q800 V600
GPA: 3.68

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22 Oct 2014, 19:23
2
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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Joined: 03 Aug 2015
Posts: 56
Concentration: Strategy, Technology
Schools: ISB '18, SPJ GMBA '17
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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
Intern
Joined: 27 Oct 2015
Posts: 19

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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
Intern
Joined: 23 Jun 2016
Posts: 14

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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.

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

Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 49271

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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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Intern
Joined: 06 Apr 2017
Posts: 29
Location: United States (OR)
Schools: Haas EWMBA '21
GMAT 1: 730 Q48 V44
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WE: Corporate Finance (Health Care)

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25 May 2017, 15:25
2
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.

Intern
Joined: 17 Sep 2017
Posts: 12
Concentration: Finance, Entrepreneurship
GMAT 1: 730 Q48 V42
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05 Oct 2017, 14:43
1
1
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
Intern
Joined: 17 Apr 2018
Posts: 9
GMAT 1: 640 Q49 V28

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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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 49271

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10 Jun 2018, 00:20
1
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 ?

Numbers on the GMAT are real by default (GMAT deals with only real numbers), so no need to consider complex roots.
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Re: M12-17 &nbs [#permalink] 10 Jun 2018, 00:20
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# M12-17

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