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Algebra - Quadratic Equations

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by shankar.ashwin » Tue Nov 01, 2011 9:26 pm
If the roots of the quadratic equation ax^2 + bx +2 = 0 are integers, how many possible values could (a,b) take given that 'a' and 'b' are integers and a>2 and b<6 ?

A) None
B) 1
C) 2
D) 8
E) More than 8
Last edited by shankar.ashwin on Wed Nov 02, 2011 3:28 am, edited 1 time in total.
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by rijul007 » Tue Nov 01, 2011 10:01 pm
i got 6 :(
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by shankar.ashwin » Tue Nov 01, 2011 10:33 pm
I got a different number aswell :P Could you post your solution?
rijul007 wrote:i got 6 :(
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by [email protected] » Tue Nov 01, 2011 10:39 pm
Please give the detailed solution
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by neelgandham » Wed Nov 02, 2011 3:03 am
shankar.ashwin wrote:If the roots of the quadratic equation ax^2 + bx +2 = 0 are real, how many possible values could (a,b) take given that 'a' and 'b' are integers and a>2 and b<6 ?

A) None
B) 1
C) 2
D) 8
E) More than 8
IMO E

Roots of ax^2 + bx +2 = 0 are real => (b^2)-(4*a*2) >0

=> b^2 > 8a

Let a = 2 (Yes, Just 2) b can be 6,5,4,-4,-5,-6,-7,-8,-9,-10.... So more than 8

Correct me if I am wrong !
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by shankar.ashwin » Wed Nov 02, 2011 3:30 am
Everyone, a small mistake in the problem. The roots are given to be integers (Not just real). Edited the post. Apologies!!
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by gmatblood » Wed Nov 02, 2011 4:05 am
IMO: B (ONE)
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by user123321 » Wed Nov 02, 2011 4:17 am
shankar.ashwin wrote:If the roots of the quadratic equation ax^2 + bx +2 = 0 are integers, how many possible values could (a,b) take given that 'a' and 'b' are integers and a>2 and b<6 ?

A) None
B) 1
C) 2
D) 8
E) More than 8
is it E?

I got 10 possibilities.

The trick here is sqrt(b^2 - 4ac) should be a perfect square and
-b+/-above value should be divisible by 2

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by GMATGuruNY » Wed Nov 02, 2011 7:49 am
shankar.ashwin wrote:If the roots of the quadratic equation ax^2 + bx +2 = 0 are integers, how many possible values could (a,b) take given that 'a' and 'b' are integers and a>2 and b<6 ?

A) None
B) 1
C) 2
D) 8
E) More than 8
Given ax²+bx+c=0, the product of the roots = c/a and the sum of the roots = -b/a.
Since the roots are integers, their product (c/a) and their sum (-b/a) must also be integers.

If ax²+bx+2=0, then c=2, implying that the product of the roots = 2/a.
Since a≥2, and 2/a is an integer, we know that a=2 and that the product of the roots = c/a = 2/2 = 1.

Since the roots are integers, there are only two possible cases:
Both roots are 1 (so that their product is 1*1 = 1 and their sum is 1+1 = 2).
Both roots are -1 (so that their product is -1 * -1 = 1 and their sum is -1 + -1 = -2).

If the sum of the roots is 2, then -b/2 = 2 and b=-4.
If the sum of the roots is -2, then -b/2 = -2 and b=4.

Thus, there are two possibilities for (a,b):
(2,4) and (2,-4).

The correct answer is C.
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by shankar.ashwin » Wed Nov 02, 2011 7:58 am
Great solution Mitch! Appreciate it.
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