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

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

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New post 09 Mar 2016, 11:53
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A
B
C
D
E

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

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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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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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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)

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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] 20 Sep 2017, 05:16
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