Here comes our next question
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
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.
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
It is given in the question that a>0
ie q-p>=0 (since p and q are integers)
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.
Statement 2 alone is not sufficient..........