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# M12 Q17

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Manager
Status: A continuous journey of self-improvement is essential for every person -Socrates
Joined: 02 Jan 2011
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Kudos [?]: 3 [0], given: 14

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02 Mar 2011, 20:41
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.

Please explain that how we have assumed $$x^2 \geq 0$$ at the start of the problem and what is the theory behind this assumption. I have also solved with the long method as mentioned by Bunnel.

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02 Mar 2011, 21:12
x^2 >= 0 as square of a real number is never negative.
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Senior Manager
Joined: 01 Nov 2010
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Kudos [?]: 73 [1] , given: 44

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03 Mar 2011, 22:52
1
KUDOS
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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Manager
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Kudos [?]: 5 [0], given: 104

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05 Mar 2012, 09:33
Got the question wrong, just went too quickly and didn't really think it through. Reading the explanations makes complete sense though.
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Joined: 15 Apr 2011
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Kudos [?]: 5 [0], given: 8

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05 Mar 2012, 23:58
Took me a while but in for (A)...
Although... I feel that this might be out of scope question for GMAT... Right?
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Kudos [?]: 48 [1] , given: 16

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26 Apr 2012, 05:06
1
KUDOS
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: $$x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4$$ --> $$2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2$$. At this point you should square again --> $$4x^4+12x^2+8=1-4x^2+4x^4$$ --> $$16x^2=-7$$, 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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Kudos [?]: 6 [1] , given: 15

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06 Mar 2013, 22:48
1
KUDOS
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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08 Mar 2013, 00:02
Tricky question. After 2 minutes I had to pick answer randomly and got this question wrong... I saw because x^2 >= 0, so the left side is always > 2. Very nice question.
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27 Apr 2014, 00:58
x^2+1>=1 and x^2+2>=2 so the left side >= sqrt(1)+sqrt(2) >2sqrt(1) = 2

Hence the equation has no root.
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05 May 2014, 13:01
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.

The solution explained by you is just awesome.
Though I solved the problem but I use lot of algebra and took much time to solve the problem.
Nice explanation
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Re: M12 Q17   [#permalink] 05 May 2014, 13:01

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# M12 Q17

Moderator: Bunuel

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