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Re: x^2 + bx + 72 = 0 has two distinct integer roots; how many values are [#permalink]
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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.

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Re: x^2 + bx + 72 = 0 has two distinct integer roots; how many values are [#permalink]
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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.

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x^2 + bx + 72 = 0 has two distinct integer roots; how many values are [#permalink]
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Kem12 wrote:
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.

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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.

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Re: x^2 + bx + 72 = 0 has two distinct integer roots; how many values are [#permalink]
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Bunuel wrote:
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


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
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Re: x^2 + bx + 72 = 0 has two distinct integer roots; how many values are [#permalink]
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Re: x^2 + bx + 72 = 0 has two distinct integer roots; how many values are [#permalink]
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