For how many values of a greater than 0 (A0), both the roots of ax^2

Question

For how many values of  a  greater than 0 ( a > 0 ), both the roots of  ax^2 - (a+1)x + (a-2) = 0 , are greater than 3?

A. 0
B. 1
C. 2
D. 4
E. 5

chetan2u · Accepted Answer

If ax^2+bx+c=0, then SUM of roots = \frac{-b}{a} and PRODUCT of roots = \frac{c}{a} Here, a=a, b=-(a+1), and c=a-2. Thus, PRODUCT = \frac{a-2}{a}>3*3............9a0, so no values will fit in for PRODUCT. A

kungfury42 · Answer

I don't know if I got really lucky on this one or if this is actually smart work, but here's what I did:

If both the roots are greater than 3, then their sum is greater than 6

therefore, (a+1)/a > 6 or a(5a-1)<0 or a is (0, 1/5)

Now here is where this gets really trippy. This is a range, and a range contains infinitely many values which satisfy. However, we have been asked distinct values of a here, which means we would have to check other conditions such as Discriminant>=0 or Minima at>=3 or Product>=9 and then arrive at an intersection of all these and see how many values of a will satisfy.

But the catch is that we will again get a range of a as a resultant and there will be infinitely many values in that range. In order to be able to pinpoint any distinct values for a, it has to be that only the edge cases a=0 and a=1/5 result from that intersection, but we know that both a=0 and a=1/5 are not a part of the solution set, and we have practically no way of defining the edge case since there is an open bracket on both 0 and 1/5, it is practically equivalent to asking what comes immediately after 0 and check for that? We can zoom infinitely on the number line and yet not arrive at that figure.

Therefore, we MUST have 0 distinct solutions for a that satisfy.

Bunuel   chetan2u  would really appreciate if you could review this and share your thoughts too. Definitely I know that there is a standard way of doing this question as well, but what I've done though seems about right to me, would like for you to review.

Posted from my mobile device

chetan2u · Answer

you are wrong in colored portion. \frac{a+1}{a}>6.......a+1>6a.........5a<1.....a<\frac{1}{5} . Also a>0. Also, product of roots >9.... and you can further work on it

kungfury42 · Answer

Hi  chetan2u , thanks for your reply. I'm not sure which step in the below is wrong, also as both of us are getting the same solution set (0, 1/5) basis that inequality. Here's what I did step by step in the red

\frac{(a+1)}{a}>6 
=  \frac{(a+1)}{a}-6>0 
=  \frac{(a+1-6a)}{a}>0 
=  \frac{(1-5a)}{a}>0 
Multiplying both sides by  a^2 
=  a*(1-5a)>0 
Multiplying both sides by  -1 
=  a*(5a-1)<0

The above inequality gets satisfied for all a between (0, 1/5)

Could you please help point out?

chetan2u · Answer

You are correct but it can be done faster by knowing a>0.
So you can cross multiply..... \frac{(a+1)}{a}>6........a+1>6a.........5a<1

kungfury42 · Answer

Admittedly, this is a faster way since a>0.

But using either way once I do derive that a is (0, 1/5) from my 1st case, is the rest of the hypothesis in my original post correct? Would really appreciate your review on that.

Posted from my mobile device

Gmat860sanskar · Answer

I'm unable to understand this problem. someone help me please.

babataer1 · Answer

1. Sum of the roots:  -b/a = (a+1)/a > 3+3 = 6
a+1 > 6a 
a < 1/5 
--> so a must be between 0 and 1/5 exclusive

2. Product of roots:   c/a = (a-2)/a > 3*3 = 9 
a-2 > 9a 
-2/8 > a but we said a>0 --> no solution

SM15 · Answer

Root 1 + Root 2 >6 Root 1 + root 2 = -b/a = ( a+1)/a >6 a<1/5 Root 1 * Root 2 - c/a > 9 a-2/a>9 a<-1/4 I'm asked greater than 3, Hence 0 values.

For how many values of a greater than 0 (A>0), both the roots of ax^2

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