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M12 Q17

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M12 Q17 [#permalink]  09 Aug 2008, 06:21
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How many roots does this equation have?

$$sqrt(x^2+1) + sqrt(x^2+2) = 2$$

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

[Reveal] Spoiler: OA
A

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Re: M12 Q17 [#permalink]  09 Aug 2008, 08:09
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Square original equation
x^2+1 + x^2+2 + 2sqrt[(x^2+1)(x^2+2)] = 4
=> 2sqrt[(x^2+1)(x^2+2)] = 1 - 2x^2

Square again
4(x^4+3x^2+2) = 1 + 4x^4 - 4x^2
=>8x^2 = -7
=> x^2 = -7/8

As x^2 is -ve, so this equation cannot have solution for any real value of x. Although complex numbers solutions can be obtained, these are not part of regular GMAT.

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Re: M12 Q17 [#permalink]  09 Aug 2008, 18:15
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abhijit_sen wrote:
Square original equation
x^2+1 + x^2+2 + 2sqrt[(x^2+1)(x^2+2)] = 4
=> 2sqrt[(x^2+1)(x^2+2)] = 1 - 2x^2

Square again
4(x^4+3x^2+2) = 1 + 4x^4 - 4x^2
=>8x^2 = -7
=> x^2 = -7/8

As x^2 is -ve, so this equation cannot have solution for any real value of x. Although complex numbers solutions can be obtained, these are not part of regular GMAT.

thanks Abhijit,
technically the language of the question is worng, should be changed to
"How many real roots does this equation have"
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Re: M12 Q17 [#permalink]  12 Aug 2008, 08:43
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Expert's post
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.
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Re: M12 Q17 [#permalink]  14 Aug 2008, 12:09
1
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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.

Agreed .. good point.
+1 for you.

Durgesh,
Question wording need to be changed to real roots to make it clear.
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Re: M12 Q17 [#permalink]  21 Aug 2008, 01:42
Thanks guys. We'll change the wording of the question.

+1 for everyone.
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Re: M12 Q17 [#permalink]  25 Feb 2010, 06:39
Quote:
Square original equation
x^2+1 + x^2+2 + 2sqrt[(x^2+1)(x^2+2)] = 4
=> 2sqrt[(x^2+1)(x^2+2)] = 1 - 2x^2

Square again
4(x^4+3x^2+2) = 1 + 4x^4 - 4x^2
=>8x^2 = -7 should be 16x^2=-7
=> x^2 = -7/8 then x^2=-7/16

As x^2 is -ve, so this equation cannot have solution for any real value of x. Although complex numbers solutions can be obtained, these are not part of regular GMAT.

But even answer is right, A
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Re: M12 Q17 [#permalink]  25 Feb 2010, 09:42
IanStewart wrote:

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.

sorry but i didnt understand this can anyone explain.
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Re: M12 Q17 [#permalink]  25 Feb 2010, 10:11
2
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Expert's post
hrish88 wrote:
IanStewart wrote:

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.

sorry but i didnt understand this can anyone explain.

$$x^2 \geq 0$$, means that the lowest value of LHS is when $$x=0$$ --> $$\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt2=2.4>RHS=2$$, hence as the lowest possible value of LHS is still greater than RHS, the equation has no real roots.

Hope it's clear.
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Re: M12 Q17 [#permalink]  25 Feb 2010, 10:58
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?
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Re: M12 Q17 [#permalink]  25 Feb 2010, 11:40
4
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Expert's post
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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Re: M12 Q17 [#permalink]  25 Feb 2010, 13:12
Hrish88,

Pardon my ignorance but what is RHS and LHS?

hrish88 wrote:
Bunuel wrote:
$$x^2 \geq 0$$, means that the lowest value of LHS is when $$x=0$$ --> $$\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt2=2.4>RHS=2$$, hence as the lowest possible value of LHS is still greater than RHS, the equation has no real roots.

Hope it's clear.

Thanks Bunuel.

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Re: M12 Q17 [#permalink]  25 Feb 2010, 13:15
1
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TheSituation wrote:
Hope it's clear

Abudantly clear...thanks. +1

Just curious, have you written the GMAT yet? If so, how did you score? If you say anything less than 790 I won't believe you..

Thanks again!!!
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Re: M12 Q17 [#permalink]  25 Feb 2010, 13:18
rbriscoe wrote:
Hrish88,

Pardon my ignorance but what is RHS and LHS?

Ha, finally a question I can help out on!!

RHS and LHS are referring to right-hand side and left-hand side of the equation.
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Re: M12 Q17 [#permalink]  25 Feb 2010, 13:21
RHS means Right Hand Side and LHS means Left Hand Side of the given equation.

here RHS is 2 and LHS is $$\sqrt{{x^2+1}} + \sqrt{{x^2+2}}$$
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Re: M12 Q17 [#permalink]  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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Re: M12 Q17 [#permalink]  25 Feb 2010, 14:05
1
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Expert's post
rbriscoe wrote:
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.

Frankly speaking I solved this question in different manner, as shown in my first post in this thread. The above solution is also valid, however it's quite time consuming.

We have: $$\sqrt{x^2+1}+\sqrt{x^2+2}=2$$. Now, if you choose to square both sides, then when squaring LHS $$\sqrt{x^2+1}+\sqrt{x^2+2}$$, you should apply the formula: $$(a+b)^2=a^2+2ab+b^2$$ --> $$(\sqrt{x^2+1})^2+2{(\sqrt{x^2+1})}{(\sqrt{x^2+2})}+(\sqrt{x^2+2})^2=x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2$$. When squaring RHS yuo'll get $$2^2=4$$.

So you'll get: $$x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2=4$$, rearrange --> $$2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2$$. At this point you should square again --> $$(2\sqrt{(x^2+1)(x^2 +2)})^2=(1-2x^2)^2$$ --> $$4(x^2+1)(x^2 +2)=1-4x^2+4x^4$$ --> $$4x^4+12x^2+8=1-4x^2+4x^4$$, rearrange again, $$4x^4$$ will cancel out --> $$16x^2=-7$$ --> $$x^2=-\frac{7}{16}$$. Now, $$x^2$$ can not equal to negative number, which means that there doesn't exist an $$x$$ to satisfies the given equation. So, no roots.

Hope it's clear.
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Re: M12 Q17 [#permalink]  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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Re: M12 Q17 [#permalink]  24 Mar 2010, 03:55
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don't think we need hard algebra,
just under the square should be x^2+1>0 or x^2+2>0
so,
x^2>-1 incorrect
x^2>-2 incorrect
thus no real roots... answer is A) 0
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Re: M12 Q17 [#permalink]  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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Re: M12 Q17   [#permalink] 02 Mar 2011, 18:59

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