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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]  08 Nov 2009, 13:02
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Question Stats:

45% (02:25) correct 55% (01:55) wrong based on 85 sessions
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?
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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.

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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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WE: Supply Chain Management (Consulting)
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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
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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