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# If equation ах^2 + bх + с = 0 have two distinct roots, which of the

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If equation ах^2 + bх + с = 0 have two distinct roots, which of the  [#permalink]

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Updated on: 12 May 2017, 09:55
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00:00

Difficulty:

45% (medium)

Question Stats:

58% (01:20) correct 42% (01:22) wrong based on 123 sessions

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If equation ах^2 + bх + с = 0 have two distinct roots, which of the following must be true?

I. b > 0
II. ac > 0
III. ac < 0

(A) None
(B) I only
(C) II only
(D) III only
(E) I and II only

Originally posted by banksy on 17 Mar 2011, 13:43.
Last edited by Bunuel on 12 May 2017, 09:55, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: If equation ах^2 + bх + с = 0 have two distinct roots, which of the  [#permalink]

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23 Mar 2011, 17:46
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banksy wrote:
SC) If equation ах^2 + bх + с = 0 have two distinct roots, which of the following must be true?
I. b > 0
II. ac > 0
III. ac < 0
(A) None
(B) I only
(C) II only
(D) III only
(E) I and II only

$$ax^2 + bx + c = 0$$ will have 2 distinct roots (assumed to be real since we are dealing with GMAT here) when $$b^2 - 4ac > 0$$
(Note that when $$b^2 - 4ac = 0$$, the roots are equal and when $$b^2 - 4ac < 0$$, the roots are imaginary)

So $$b^2 > 4ac$$ is the condition required.
Is it necessary that b should be positive? No. Even if b is negative, $$b^2$$, which will be positive, can be greater than 4ac.
There is no condition at all on ac. It can be positive or negative. Just that $$b^2$$ should be greater than it.
Hence we don't need any one of them to be necessarily true.
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Re: If equation ах^2 + bх + с = 0 have two distinct roots, which of the  [#permalink]

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17 Mar 2011, 14:49
banksy wrote:
SC) If equation ах^2 + bх + с = 0 have two distinct roots, which of the following must be true?
I. b > 0
II. ac > 0
III. ac < 0
(A) None
(B) I only
(C) II only
(D) III only
(E) I and II only

we know that if a,b,c is +ve then both roots of x = -ve
roots of x = (-b+\sqrt{b^2 - 4ac})/2a and x = (-b-\sqrt{b^2 - 4ac})/2a

both of these roots will -ve if b>0. \sqrt{b^2 - 4ac} will have to be +ve for it to be a real number. I dont think we can say if ac needs to be +ve or -ve. I will go with B.

if we cannot assume a,b,c to be +ve. then I'll go with A.
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Re: If equation ах^2 + bх + с = 0 have two distinct roots, which of the  [#permalink]

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17 Mar 2011, 18:28
also goes with A. the same solution with dream
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Re: If equation ах^2 + bх + с = 0 have two distinct roots, which of the  [#permalink]

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17 Mar 2011, 20:51
1
banksy wrote:
SC) If equation ах^2 + bх + с = 0 have two distinct roots, which of the following must be true?
I. b > 0
II. ac > 0
III. ac < 0
(A) None
(B) I only
(C) II only
(D) III only
(E) I and II only

For a quadratic equation to have distinct root;

$$b^2>4*a*c$$

I. What if b=0

0 > -ve
Any negative result of a*c will suffice to make the quadratic equation to have two distinct roots

Let's choose
a=1
c=-1
b=0
1x^2-1=0
x^2=1
$$x=\pm 1$$
Two roots.
Not true that b must be more than 0.

II. What if ac<0

Let's choose
a=1
c=-1
b=0
1x^2-1=0
x^2=1
$$x=\pm 1$$
Two roots.
Not true that ac must be more than 0.

III. What if ac>0

a=1
b=4
c=3

x^2+4x+3=0
x^2+3x+x+3=0
x(x+3)+(x+3)=0
(x+1)(x+3)=0
x=-1
x=-3
Two distinct root.
Not true that ac < 0 must be true.

Ans: "A"
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Re: If equation ах^2 + bх + с = 0 have two distinct roots, which of the  [#permalink]

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18 Mar 2011, 08:42
Fluke I love your lucid style. It's awesome and fantabulistic!

Posted from my mobile device
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Re: If equation ах^2 + bх + с = 0 have two distinct roots, which of the  [#permalink]

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08 May 2011, 18:39
the equation have distinct roots.

=> for an equation to have distinct roots,

b>0 need not be true . b =0 , b<0 are possible too.

II is not a must.
III is not a must.

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Re: If equation ах^2 + bх + с = 0 have two distinct roots, which of the  [#permalink]

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08 May 2011, 19:29
using (x+2) (x+4) | (x-2) (x+4) | (x+2) (x-4) we get different values for b and ac.

Hence A.
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Re: If equation ах^2 + bх + с = 0 have two distinct roots, which of the  [#permalink]

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29 Jul 2018, 05:07
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Re: If equation ах^2 + bх + с = 0 have two distinct roots, which of the &nbs [#permalink] 29 Jul 2018, 05:07
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