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Q1. How many roots does this equation have? A*x + A = B 1. A

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Senior Manager
Joined: 05 Oct 2008
Posts: 263
Q1. How many roots does this equation have? A*x + A = B 1. A [#permalink]

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02 Nov 2008, 11:46
Q1. How many roots does this equation have? A*x + A = B

1. A does not equal 0
2. B does not equal 0

Q2. How many real roots does this equation have?

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

Can anyone explain the logic in such questions? What does it mean "how many "roots"?"

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Current Student
Joined: 28 Dec 2004
Posts: 3310
Location: New York City
Schools: Wharton'11 HBS'12

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02 Nov 2008, 13:54
for the first one..we need to know if A is not zero..
therefore A it is for the first one..

for the second one, i think you have 2 roots..

the power of the polynomial determines the number of roots.
Senior Manager
Joined: 05 Oct 2008
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03 Nov 2008, 11:16
Answer to the first one is correct, but another try at the 2nd one?
Current Student
Joined: 28 Dec 2004
Posts: 3310
Location: New York City
Schools: Wharton'11 HBS'12

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03 Nov 2008, 11:22
study wrote:
Answer to the first one is correct, but another try at the 2nd one?

well its not clear how the question is laid out .. if its not 2 then its 4..
Intern
Joined: 29 Oct 2008
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03 Nov 2008, 13:51
study wrote:
Answer to the first one is correct, but another try at the 2nd one?

No real roots.

x^2 is always positive.
So min value for x^2+1 = 1 and min value for x^2+2 = 2
Hence, sqrt(1) + sqrt(2) > 2.. Dont know if any catch in here..
Current Student
Joined: 28 Dec 2004
Posts: 3310
Location: New York City
Schools: Wharton'11 HBS'12

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03 Nov 2008, 14:22
way2 wrote:
study wrote:
Answer to the first one is correct, but another try at the 2nd one?

No real roots.

x^2 is always positive.
So min value for x^2+1 = 1 and min value for x^2+2 = 2
Hence, sqrt(1) + sqrt(2) > 2.. Dont know if any catch in here..

x^2 has to real roots, x^4 has 4 real root.power of polynomial determines the number of roots.
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Joined: 23 Aug 2008
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03 Nov 2008, 21:43
3
KUDOS
study wrote:
Q1. How many roots does this equation have? A*x + A = B

1. A does not equal 0
2. B does not equal 0

Q2. How many real roots does this equation have?

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

Can anyone explain the logic in such questions? What does it mean "how many "roots"?"

My understanding is that roots are "number of possible values of x" OR "number of solutions to the equation"

Q1: Simplify to A (x+1) = B; If A != 0, there is one root: (B/A)-1, Sufficient. B is not sufficient, need to know A.

Q2: There's no real roots for this?
Minimum value is at x=0, which gives Sqrt(1) + Squrt(2) ~= 1+1.4 = 2.8
since x^2 is always positive, any -ve or +ve value of x will give a high number for this. There is probably an unreal root though, since x^2 in that case could be negative.

BTW Note that the polynomial does _not_ determine the number of roots. I'll include a link below that explains this; or simply think of x^4=0, which only has 1 root at x=0

http://www.sosmath.com/algebra/factor/fac03/fac03.html

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Re: Roots   [#permalink] 03 Nov 2008, 21:43
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Q1. How many roots does this equation have? A*x + A = B 1. A

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