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# M12 Q17

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Manager
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25 Feb 2010, 13:34
I am sooo lost on this question. I solved it the same as TheSiuation. I am do not understand your method below. Although I am sure it is pbvious to everyone but please clarify the rule, therum or postulate that you followed to solve the equation. Thanks in advacne.

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.

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26 Feb 2010, 23:30
at first glance i started substituting values and identified that x^2 cannot be greater than zero
if so then it wouldn't satisfy the equation

but i really don't what they meant as real roots and thanks bunuel for the explanation.if i had known that it is very to identify that if x^2<0 then it has no real solution
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02 Mar 2011, 18:59
I tried like this :

Let (x^2 + 1) = y

sqrt(y) + sqrt(y+1) = 2

y + y + 1 + 2(sqrt(y) * sqrt(y+1)) = 4

2y + 2(sqrt(y) * sqrt(y+1)) = 3

=> 4 * y * (y+1) = 4y^2 + 9 - 12y

=> 4y^2 + 4y = 4y^2 + 9 - 12y

=> y = 9/16

=> x^2 + 1 = 9/16 , not possible so 0 roots, answer is A.
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02 Mar 2011, 19:41
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, 20:12
x^2 >= 0 as square of a real number is never negative.
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05 Mar 2012, 08:33
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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05 Mar 2012, 22:58
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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07 Mar 2013, 23:02
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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26 Apr 2014, 23:58
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, 12:01
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, 12:01

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# M12 Q17

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