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From A we know that -(b/a)>0 , given a>0 then b must be NEG
In order both roots to be positive , discriminant must be >0
or b^2-4*a*c>0 .
A) is not suff

From B) we get (4a*c)/4*(a^2) or c/a>0 provided a>0 then c is >0
Given a+b+c>0 and b<0 then a+c>b
Substitute in equation of discriminant b with a+c and get
(a+c)^2-4ac>0 we get a^2-2ac+c^2>0
this is (a-c)^2>0 which is always positive BUT we do not know if A is not EQUAL TO C which would make (a-c)^2=0
SO E is the answer

4. The roots of a quadratic equatin ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

(1) Sum of the roots is positive.
(2) Product of the roots is positive.
IMHO C it is
1st alone
roots may be -2 and 4 sum is +ve
roots may be 2 and 4 and so on insuff
2)both +ve and both -ve insuff
both together -->from 2 we know that throots must be both +ve or -ve
but in order to have positive sum both should be +ve only

4. The roots of a quadratic equatin ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

(1) Sum of the roots is positive. (2) Product of the roots is positive.

PRIZE: Vocabulary Practice Sotware (contains 3000 words)

ax^2+bx+c=0 => x^2 +(b/a)x +(c/a)=0 since a>0, a is not 0

If this quadratic equation has two integer roots m and n, then it can be written as (x-m)(x-n)=0 where mn=c/a and -(m+n)=b/a

This is because (x-m)(x-n)=x^2-(m+n)x+mn

(1) We are told than m+n>0, which means that b/a<0 and so b<0 Also as m and n are integers m+n>=1, so |b/a|>=1 i.e |b|>=|a|=a

Summarizing, we know that a>0, b>0 and |b|=-b>=a

This means that b+a<=0 and since a+b+c>0, c>0 Thus ac>0 and so the roots have the same sign-

If the sum of the roots is positive, each must be >0 SUFFICIENT

(2) mn>0 means roots have the same sign. This doesn't tell me much.

Could they both be <0? Sure! (x+2)(x+1)=x^2+3x+2=0 has roots of -2 and -1 and a+b+c=6>0

Could they both be<0? Why not? (x-2)(x-6)=x^2-8x+12=0 has roots of 2 and 6 and a+b+c=5>0

NOT SUFFICIENT

My answer: A

Yes kevin, the answer is A.
I have a different approach...........

The roots of a quadratic equatin ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

Statement 1: Sum of the roots is positive.

ie -b/a = p where p>0
ie b= -(a x p).

Now consider the product of the roots.

Let c/a=q.

Clearly both p and q must be integers (since in the question it is given that both the roots are integers)

It is given that a+b+c>0
ie a-(axp)+(axq)>0
ie a(1-p+q)>0
It is given in the question that a>0
So (1-p+q)>0
ie q-p>-1
ie q-p>=0 (since p and q are integers)
ie q>=p
In statement 1 it is given that p>0
So clearly q>0

So from the first statement itself if sum is +ve we can conclude that product is also +ve.

5. In triangle ABC angle A is the greatest angle. D is the foot of the
perpendicular dropped on to BC from A. Is triangle ABC right-angled?
1. AD^2= BD x DC.
2. AD/DC < BD/AD.

Prize: A file on logical ability to solve the critical reasoning questions from GMAT.

5. In triangle ABC angle A is the greatest angle. D is the foot of the perpendicular dropped on to BC from A. Is triangle ABC right-angled? 1. AD^2= BD x DC. 2. AD/DC < BD/AD.

Prize: A file on logical ability to solve the critical reasoning questions from GMAT.