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

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

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New post 02 Nov 2008, 11:46
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Re: Roots [#permalink]

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New post 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.

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Re: Roots [#permalink]

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New post 03 Nov 2008, 11:16
Answer to the first one is correct, but another try at the 2nd one?

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Re: Roots [#permalink]

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New post 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..

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Re: Roots [#permalink]

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New post 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..

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Re: Roots [#permalink]

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New post 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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Re: Roots [#permalink]

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New post 03 Nov 2008, 21:43
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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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