x^2 + bx + 72 = 0 has two distinct integer roots; how many values are

Question

x^2 + bx + 72 = 0 has two distinct integer roots; how many values are possible for b?

A. 3
B. 6
C. 8
D. 12
E. 24

AneesShaik · Accepted Answer

For a quadratic equation ax^2+bx+c=0, we know that -b/a is sum of roots and c/a is product of roots. The quadratic equation here is x^2 + bx + 72 = 0, where product of roots is 72. If we find all the factors of 72, we have the answer. By prime factorization, we get 72= 2^3*3^2. We know that total factors are (3+1)*(2+1) = 12 (Reason: with 2^n, we have n+1 possibilities. n^0 to n^n. so n+1) Anees Shaik. :-D

Divyadisha · Answer

With confused head, I want to choose 'B' as the answer.

As 72 is positive either both roots are + or -

but as 'b' is +ve, both the roots carry same sign, i.e. +ve, and the factors are as below
72, 1
2, 36
4, 18
8, 9
24, 3
6, 12

Hope I have chosen the right answer.

EMPOWERgmatRichC · Answer

Hi Divyadisha,

All of the work you've shown is correct, BUT you forgot to factor in that "b" could be positive or negative.

For example, your first 'pair' of numbers: 1 and 72, could actually be -1 and -72. In those two examples, "b" would equal +73 OR -73, depending on whether the integers were positive or negative. Thus, you have to double the number of options that you've listed.

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

Kem12 · Answer

Hi  Bunuel , I started out saying that b^2-4ac>0 must be true for this equation to have 2 distinct integer roots but got stuck at b^2-288>0 so I used the above explanation to arrive at the answer however I am very curious as to why the approach did not work in this case or maybe I applied it wrongly. Thanks. Season greetings.

Posted from my mobile device

ruchik · Answer

Hi Kem the condition you have mentioned is a necessary condition for an equation to satisfy for the roots to be distinct and real. But it does not help in finding the roots of equations which the question here asks.
What i mean to say you need to find two roots of the equation which is given by  (-b+\sqrt{D})/2a  and  (-b-\sqrt{D}])/2a where D = b^2-4ac.
The necessary condition for above two roots to be distinct and real is b^2-4ac > 0

Hope it helps.

Give kudos if you like the explanation,

Kinshook · Answer

Given: x^2 + bx + 72 = 0 has two distinct integer roots; 
Asked: how many values are possible for b?

72 = 2^3*3^2
Factors of 72 = {1,2,3,4,6,8,9,12,18,24,36,72}

b is the summation of 2 factors of 72

|b| = 1 + 72 = 73; b = 73 or -73
|b| = 2 + 36 = 38; b = 38 or -38
|b| = 3 + 24 = 25; b = 25 or -25
|b| = 4 + 18 = 22; b = 22 or -22
|b| = 6 + 12 = 18; b = 18 or -18
|b| = 8 + 9 = 17; b = 17 or - 17

Total values of b = 6*2 = 12

IMO D

x^2 + bx + 72 = 0 has two distinct integer roots; how many values are

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

x^2 + bx + 72 = 0 has two distinct integer roots; how many values are

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards