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

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Math Expert
Joined: 02 Sep 2009
Posts: 55670
x^2 + bx + 72 = 0 has two distinct integer roots; how many values are  [#permalink]

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09 Mar 2016, 12:53
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75% (hard)

Question Stats:

52% (02:02) correct 48% (02:13) wrong based on 201 sessions

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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

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

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24 Aug 2016, 08:03
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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

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

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09 Mar 2016, 20:49
1
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.
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Re: x^2 + bx + 72 = 0 has two distinct integer roots; how many values are  [#permalink]

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10 Mar 2016, 21:33
3

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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31 Dec 2018, 02:49
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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31 Dec 2018, 22:25
1
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.

Posted from my mobile device

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,
x^2 + bx + 72 = 0 has two distinct integer roots; how many values are   [#permalink] 31 Dec 2018, 22:25
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