M12-17

Awesome approach...I tried the same thing...squared on both sides and then got spiraled within the roots...

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.

Answer A

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.

Bunuel VeritasKarishma,

Can you please let me know the concept involved in your approach?

In case of LHS = RHS or LHS<RHS, then what will be inferred?

Regards!!

'Roots of the equation' means the values of x for which the equation will be true.

\(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\)

The point here is that the equation CANNOT be true for any values of x.

When I look at the equation, I don't feel like squaring because even after squaring, I will need to square again to get rid of the square root. So instead I notice the 2 on the RHS and wonder what values the square roots on LHS can take. Since square roots will be positive only, I wonder whether each can be 1. Then I notice that inside the square root, I have x^2 which must be 0 or positive and 1. So the first square root can be 1. I notice the second square root and see that x^2 must be 0 or positive and there is a 2. So the second square root must be at least 1.4 (sqrt2).
Since everything is all positive, I see that LHS must be at least 2.4 (when x = 0). Then can LHS be 2? No. It cannot be for ANY value of x. So for NO value of x will the equation be satisfied.
Hence number of roots = 0.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Amazing and beautiful question indeed!

The equation is indeed jarring and scary, and is tempting to try to solve by squaring or taking roots (I know I did, tried for almost 3 minutes )

But after reading again, I find a 30 seconds solution by realizing that this is not an algebra but a number property/theory problem

reiterating the question
\(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\)

Let's think for a sec. Okay, so we need two roots of a number that equals 2. Fair enough.

Also notice that both of the terms have an \(x^2\) term, which is always positive or zero, \(x>=0\). Up to this point, they're not helping at all.

But wait! Each of the terms are always greater than 1!
For the first term \(\sqrt{x^2 + 1} \), min value is 1 when \(x^2 = 0\)
For the second term \(\sqrt{x^2 + 2} \), min value is \(\sqrt{2} \approx 1.4 \) when \(x^2 = 0\).

You dont actually need the approximate number of 1.4 to solve this problem, only that obviously \(\sqrt{2} > \sqrt{1} \) or that \(\sqrt{2}\) MUST BE greater than 1

Now, that since the first square root term is 1 and the second is greater than 1, adding both CAN NOT BE EQUAL to 2, thus it doesn't have any solution in real numbers, hence A

I think this is a high-quality question and I agree with explanation.

\(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) ... (1)

Observe that:
\(\{sqrt{x^2 + 1}}^2 - \{sqrt{x^2 + 2}}^2 = x^2 +2 - (x^2 - 1) = 1\)

Dividing this by the give equation and using a²-b²=(a+b)(a-b), we have:

\(\sqrt{x^2 + 1} - \sqrt{x^2 + 2} = 1/2\) ... (2)

Add 1 and 2:
2 * sqrt{x^2 + 2} = 2 + 1/2 = 5/2
=> sqrt{x^2 + 2} = 5/4
=> x² + 2 = 25/16
=> x² = 25/16 - 2 < 0
Thus, there is no real solution

Answer A

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