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# The roots of the equation Ax^3 + Bx^2 + C^x + D = 0, where A

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The roots of the equation Ax^3 + Bx^2 + C^x + D = 0, where A [#permalink]  03 Sep 2009, 23:48
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The roots of the equation Ax^3 + Bx^2 + C^x + D = 0, where A > 0, are in A.P. Find all possible roots of the equation Bx^2 + C^x + D = 0.

I. One of the roots of the equation Ax^3 + Bx^2 + Cx + D = 0 is 0.
II. C < 0.
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Re: DS - Tough [#permalink]  04 Sep 2009, 01:32
GODSPEED wrote:
are in A.P.

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Re: DS - Tough [#permalink]  04 Sep 2009, 02:09
Does A.P. mean arithmetic progression?
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Re: DS - Tough [#permalink]  04 Sep 2009, 09:40
IMO A....well I am unable to understand use of A.P. .here..

Stmt1 gives D = 0.....
so Bx^2 + C^x + D = 0 become Bx^2 + C^x=0
solving two roots will be 0 , -C/B
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Re: DS - Tough [#permalink]  04 Sep 2009, 12:35
CasperMonday wrote:
Does A.P. mean arithmetic progression?

Yeah! I'm sorry for the confusion. A.P. means Arithmetic Progression
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Re: DS - Tough [#permalink]  04 Sep 2009, 12:57
age wrote:
IMO A....well I am unable to understand use of A.P. .here..

Stmt1 gives D = 0.....
so Bx^2 + C^x + D = 0 become Bx^2 + C^x=0
solving two roots will be 0 , -C/B

here, C and B are unknown..question is asking to find out the values of C and B. i think S! is insufficient as well..

I would go with E,
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Re: DS - Tough [#permalink]  04 Sep 2009, 17:51
The roots of the equation Ax^3 + Bx^2 + C^x + D = 0, where A > 0, are in A.P. Find all possible roots of the equation Bx^2 + C^x + D = 0.

I. One of the roots of the equation Ax^3 + Bx^2 + Cx + D = 0 is 0.
II. C < 0.

Shouldn't it be Ax^3+Bx^2+Cx+D=0, where A>0, are in A.P. Find all possible roots of the equation Bx^2+Cx+D=0.

If so the answer is C
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Re: DS - Tough [#permalink]  04 Sep 2009, 18:42
Bunuel wrote:
The roots of the equation Ax^3 + Bx^2 + C^x + D = 0, where A > 0, are in A.P. Find all possible roots of the equation Bx^2 + C^x + D = 0.

I. One of the roots of the equation Ax^3 + Bx^2 + Cx + D = 0 is 0.
II. C < 0.

Shouldn't it be Ax^3+Bx^2+Cx+D=0, where A>0, are in A.P. Find all possible roots of the equation Bx^2+Cx+D=0.

If so the answer is C

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Re: DS - Tough [#permalink]  04 Sep 2009, 19:38
2
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gmate2010 wrote:
Bunuel wrote:
The roots of the equation Ax^3 + Bx^2 + C^x + D = 0, where A > 0, are in A.P. Find all possible roots of the equation Bx^2 + C^x + D = 0.

I. One of the roots of the equation Ax^3 + Bx^2 + Cx + D = 0 is 0.
II. C < 0.

Shouldn't it be Ax^3+Bx^2+Cx+D=0, where A>0, are in A.P. Find all possible roots of the equation Bx^2+Cx+D=0.

If so the answer is C

O.K. It was not easy for me but I have finally got the answer. (Well hope so)

(1) One of the roots of the equation Ax^3+Bx^2+Cx+D=0 is 0 --> x_1=0 --> d=0

We'll have x(ax^2+bx+c)=0 one solution is x_1=0, another two: x_2 and x_3 will be the roots of (ax^2+bx+c)=0;

We know that for the roots of quadratic equation ax^2+bx+c=0 the sum of roots equals to \frac{b}{a}, so x_2+x_3=\frac{b}{a} and the product of the roots equals to \frac{c}{a}, so x_2*x_3=\frac{c}{a};

As far as we know that x_1, x_2, x_3 are in A.P. and one of them is 0, so either A. x_2=-x_3 meaning they are 1st and 3rd terms of A.P, therefore x_2+x_3=\frac{b}{a}=0 x_2*x_3=\frac{c}{a}<0 or B. they are both positive or both negative, therefore x_2*x_3=\frac{c}{a}>0

Not sufficient.

(2) c<0, not sufficient by itself.

(1)+(2) x_2*x_3=\frac{c}{a}<0 means that they are 1st and 3rd terms in A.P. The middle term is 0, so x_2+x_3=\frac{b}{a}=0 --> b=0. So we have b=0 and d=0 --> the equation we should find roots in will be cx=0, c<0, x=0. One solution x=0.

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Re: DS - Tough [#permalink]  04 Sep 2009, 19:49
1
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Bunuel wrote:
gmate2010 wrote:
Bunuel wrote:
The roots of the equation Ax^3 + Bx^2 + C^x + D = 0, where A > 0, are in A.P. Find all possible roots of the equation Bx^2 + C^x + D = 0.

I. One of the roots of the equation Ax^3 + Bx^2 + Cx + D = 0 is 0.
II. C < 0.

Shouldn't it be Ax^3+Bx^2+Cx+D=0, where A>0, are in A.P. Find all possible roots of the equation Bx^2+Cx+D=0.

If so the answer is C

O.K. It was not easy for me but I have finally got the answer. (Well hope so)

(1) One of the roots of the equation Ax^3+Bx^2+Cx+D=0 is 0 => x1=0 d=0
We'll have x(ax^2+bx+c)=0 one solution x1=0, another two: x2 and x3 will be the roots of (ax^2+bx+c)=0
We know that for the roots of ^2 equation x2+x3=b/a and x2*x3=c/a
As far as we know that x1,x2, x3 are in A.P. and one of them is 0, so either A. x2=-x3 meaning they are 1st and 3rd terms of A.P, therefore x2+x3=b/a=0 x2*x3=c/a<0 or B. they are both positive or both negative, therefore x2*x3=c/a>0
insufficient

(2) c<0 insufficient itself, but
(1)+(2)
x2*x2=c/a<0 means they are 1st and 3rd terms in A.P. The middle term is 0, so x2+x3=b/a=0 => b=0

So we have
b=0 and d=0 => the equation we should find roots in will be cx=0 c<0 x=0. One solution x=0.

excellent!!! kudos to you..i did not read the full explanation, but i got to know what i missed...Thanks a ton..
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Re: DS - Tough [#permalink]  29 Sep 2009, 17:19
Where did you guys find this question. Blew my brains away. I have no freaking clue how to solve this, even after reading the explanations.
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Re: DS - Tough [#permalink]  29 Sep 2009, 18:39
nice explanation dude
Re: DS - Tough   [#permalink] 29 Sep 2009, 18:39
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