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If the equation x^3 - ax^2 +bx -a = 0 has three real roots

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If the equation x^3 - ax^2 +bx -a = 0 has three real roots [#permalink] New post 08 Nov 2009, 13:02
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If the equation x^3 - ax^2 +bx -a = 0 has three real roots then the following id true

A. a=11
B. a not equal to 1
C. b = 1
D. b not equal to 1


How to solve equations with power 3?
Any specific tips to handle questions with such equations?
[Reveal] Spoiler: OA
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Re: Quadratic eq....roots......Help! [#permalink] New post 08 Nov 2009, 15:05
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papillon86 wrote:
If the equation x^3 - ax^2 +bx -a = 0 has three real roots then the following id true

a) a=11
b) a not equal to 1
c) b = 1
d) b not equal to 1

How to solve equations with power 3?
Any specific tips to handle questions with such equations?


I don't think we need to know how to solve cubic equation for GMAT.

For above question:

x^3 - ax^2 +bx -a = 0

x^2*(x-a)+(bx-a)=0

We are told that this equation has THREE roots. If look closer we notice that when b=1 we can write the equation as:

x^2*(x-a)+(x-a)=0 --> (x^2+1)*(x-a)=0 and this equation has only ONE root x=a.

Hence given equation to have three roots b must not equal to 1.

Answer: D.
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Re: Quadratic eq....roots......Help! [#permalink] New post 11 May 2011, 22:23
assuming b= 1

x^2(x-a)+ 1(x-a) = (x^2+1) * (x-a)

x^2 != -1 meaning b != 1.

Hence D.
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Re: If the equation x^3 - ax^2 +bx -a = 0 has three real roots [#permalink] New post 27 Nov 2013, 00:50
Good question.

The point here is to reduce the equation to x2(x-a)+bx-a=0.

Now if b=1, the equation becomes (x2+1)(x-a) = 0

The above equation has one real and two imaginary roots and hence does not satisfy the condition given in the quesiton(3 real roots)

Therefore b cannot be equal to 1.

Thanks.
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Re: If the equation x^3 - ax^2 +bx -a = 0 has three real roots   [#permalink] 27 Nov 2013, 00:50
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