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.


Answer: A
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Awkward question. I was trying to square both site and ultimately got stuck somewhere.
Nice approach to this ques.

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

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

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

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.

The correct answer is A

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

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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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

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

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

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

let \({x^2 + 1}\) = y

\(\sqrt{y} + \sqrt{y + 1} = 2\)-------------------------------(1)

squaring both side

\({(\sqrt{y} + \sqrt{y + 1} )}^2\) = y + (y+1) + \(2(\sqrt{y} * \sqrt{y + 1} )\) = 4

2y+1 + \(2(\sqrt{y} * \sqrt{y + 1} )\) = 4

2y + \(2(\sqrt{y} * \sqrt{y + 1} )\) = 3

2 (y + \((\sqrt{y} * \sqrt{y + 1} )\) ) = 3

2 \(\sqrt{y}\) ( \(\sqrt{y}+ \sqrt{y + 1}\) ) =3

2 \(\sqrt{y}\) (2) =3-------------------------------------------( from 1)

4 \(\sqrt{y}\) =3

\(\sqrt{y}\) =\(\frac{3}{4}\)

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

\(\sqrt{{x^2 + 1}}\) =\(\frac{3}{4}\)

\({x^2 + 1}\) = \((\frac{3}{4})^2\)

\({x^2 + 1}\) = \(\frac{9}{16}\)

\({x^2 }\) =\(\frac{9}{16} -1\)

\({x^2 }\) = -ve value

so no Solution for X. so ans is A

I think this is a high-quality question and I agree with explanation.
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I got this x^2= -ve value

So equation has 2 complex roots ? Bunuel

All numbers on the GMAT are real numbers by default. So, we can ignore complex roots/numbers altogether.
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Hello Bunuel, Can you please check my approach?

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

Let x^2+1=a

the expressions under the roots now become

\(\sqrt{a} + \sqrt{a+1} = 2\) (Note X^2+2 can be written as (x^2+1)+1) )

Squaring both sides of \(\sqrt{a} + \sqrt{a+1} = 2\)

We get a+a+1=4
\(\ a=[3][/2]\)


Substituting value of a in x^2+1= 3/2

now x^2+2=5/2

Sum of square root of 3/2 and square root of 5/2 is greater than 2.
therefore ans A
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I took an alternative approach. Bunuel do you have any thoughts?

1.) eliminate the square root signs by squaring both sides of the equation to get: \((x^2+1)\) + \((x^2+2)\) = 4
2.) combine terms: \(2x^2 + 3\) = 4
3.) simplify: \(2x^2 = 1\) --> \(x^2 = \frac{1}{2}\)
4.) plug "\(x^2 = \frac{1}{2}\)" back in to the equation
5.) \(\sqrt{\frac{1}{2} + 1}\) + \(\sqrt{\frac{1}{2} + 2}\) = 2
6.) apply logic... the square root of any number greater than 1 will yield a number greater than 1
6a.) the square root of 1.5 will be greater than 1
6b.) the square root of 2.5 will be greater than 1 (more specifically, it will be greater than 1.4 since the square root of 2 is roughly 1.4)
7.) you are now left with the equation: (>1) + (>1) = 2... which is false, thus there are no values of X which solve the equation (Answer A)

As always, please feel free to share any questions, comments, and/or concerns.