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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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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

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.

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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Please +1 KUDO if my post helps. Thank you.

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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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