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m12#17

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m12#17 [#permalink] New post 18 May 2011, 18:15
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How many real roots does this equation have?

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

a. 0
b. 1
c. 2
d. 3
e. 4


you can solve the question if you wish, my question is how do you arrive to answer to this and other these type of questions.
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Re: m12#17 [#permalink] New post 18 May 2011, 18:55
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This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.
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Re: m12#17 [#permalink] New post 26 Jun 2012, 00:46
gmat1220 wrote:
This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.


I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help :)

Thanks,
Diana
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Re: m12#17 [#permalink] New post 26 Jun 2012, 00:58
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dianamao wrote:
gmat1220 wrote:
This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.


I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help :)

Thanks,
Diana


How did you get: 4(x^2+1)(x^2+2)+8?

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\geq 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 + \sqrt2\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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Re: m12#17 [#permalink] New post 26 Jun 2012, 01:08
Bunuel wrote:
dianamao wrote:
gmat1220 wrote:
This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.


I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help :)

Thanks,
Diana


How did you get: 4(x^2+1)(x^2+2)+8?

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\geq 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 + \sqrt2\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.


First, I squared both sides of the equation. So I got something like (x^2+1)+2sqrt((x^2+1)(x^2+2))+(x^2+2)=4
Then I subbed relevant numbers into the discriminant formula which turns into 4(x^2+1)(x^2+2)+8. Now here, I'm thinking that 4(x^2+1)(x^2+2)+8 is definitely greater than zero, and if discriminant is greater than 0, then there are two roots, no?

P.S. I totally get your reasoning, it's just that I don't undersand why I can't reach that same answer using the discriminant rule, unless I'm not doing something right.

Thanks heaps!
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Re: m12#17 [#permalink] New post 26 Jun 2012, 01:31
Expert's post
dianamao wrote:
First, I squared both sides of the equation. So I got something like (x^2+1)+2sqrt((x^2+1)(x^2+2))+(x^2+2)=4
Then I subbed relevant numbers into the discriminant formula which turns into 4(x^2+1)(x^2+2)+8. Now here, I'm thinking that 4(x^2+1)(x^2+2)+8 is definitely greater than zero, and if discriminant is greater than 0, then there are two roots, no?


P.S. I totally get your reasoning, it's just that I don't undersand why I can't reach that same answer using the discriminant rule, unless I'm not doing something right.

Thanks heaps!


In the above reasoning there no quadratics shown. So, what is the quadratics you find the discriminant of? How did you get this quadratics?
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: m12#17 [#permalink] New post 26 Jun 2012, 01:38
Bunuel Oh! that's some really sensible work! thanks! great job!
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Re: m12#17 [#permalink] New post 14 Feb 2013, 05:34
Bunuel wrote:
dianamao wrote:
First, I squared both sides of the equation. So I got something like (x^2+1)+2sqrt((x^2+1)(x^2+2))+(x^2+2)=4
Then I subbed relevant numbers into the discriminant formula which turns into 4(x^2+1)(x^2+2)+8. Now here, I'm thinking that 4(x^2+1)(x^2+2)+8 is definitely greater than zero, and if discriminant is greater than 0, then there are two roots, no?


P.S. I totally get your reasoning, it's just that I don't undersand why I can't reach that same answer using the discriminant rule, unless I'm not doing something right.

Thanks heaps!


In the above reasoning there no quadratics shown. So, what is the quadratics you find the discriminant of? How did you get this quadratics?


bro bunuel,

I did the same way .. squared.. then squared again..

in the end I got, 16 * square(x)= - 7
and now there can never be square root of a negative number , hence no roots.. and hence zero..
hope this is correct. pls let me know
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Re: m12#17 [#permalink] New post 16 Feb 2013, 12:14
Bunuel wrote:
dianamao wrote:
gmat1220 wrote:
This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.


I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help :)

Thanks,
Diana


How did you get: 4(x^2+1)(x^2+2)+8?

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\geq 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 + \sqrt2\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.


Bunuel, Can you pls help me with the explanation:-X^2>=0, how did you get that ...Did you assume that since RHS is a positive term hence the LHS must be equal to 0...that implies x^2 must be greater than or equal to zero.

Also , can you pls post links to some the practice questions of above type...or the method used to determine the number of roots in the above cases.

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Re: m12#17   [#permalink] 16 Feb 2013, 12:14
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