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by liferocks » Tue May 11, 2010 3:46 am
let the second root be -2n where n is positive integer

so equation is (x-3)(x+2n)=0
or x^2-(3-2n)x-6n=0

only d and e have the constant with form -6n

if -6n=-12...n=2..so 3-2n=-1..equation x^2+x-12

if -6n=-18..n=3..so 3-2n=-2..equation x^2+3x-18

Ans option D
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by kstv » Tue May 11, 2010 3:48 am
since the factors are 3 and even integer = 2a

(x-3)(x+2a) = x² -x(3-2a) - 6a

so 6a can be 12 or 18

a = 2 or 3

-x (3-2a) will be x when a = 2 and 3x when a = 3

should be enough.
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by sanju09 » Tue May 11, 2010 5:03 am
pradeepkaushal9518 wrote:if 3 is one root of an equation and the second root is a negative even interger, which of the following expression might represent the equation?
a. x2-8x+15
b.x2-9x+18
c.x2-3x-10
d.x2+x-12
e.x2-3x-18
The wordings must have been "if 3 is one root of a quadratic equation..." here.

Please recollect that if α and β are the two roots of a quadratic equation of the form A x^2 + B x + C = 0, with A = 1, as in all the choices given, then, (α + β) = -B and α β = C. If α = 3 and β, some negative even integer, say -2, then we cannot concretely predict about the sign of B, which must be an odd integer.

Now, see through the choices

(A) and (B) are out for their C's (+15 and +18) not being negative

(C) is out as its C (-10) doesn't have a 3 as factor

(D) and (E) are the tough contenders left, but (E) fails the f (3) test,

[spoiler]which (D) wins[/spoiler]
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by debmalya_dutta » Tue May 11, 2010 1:05 pm
product of the roots of a quadratic equation ax^2+bx+c = c/a
sum of the roots of a quadratic equation ax^2+bx+c = -b/a

here one of the roots is 3 and the other root is negative & even means that c has to be negative and even
So , on first glance only options D and E are worth evaluating

in D , sum of roots = -1 => 3+ other root = -1 => other root = -4

So , D is the answer

You will see in option E
sum of roots = 3
=> 3+ other root = 3 => other root = 0 . but it is stated that the other root is negative & even . so this cannot be the equation
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by stephen@knewton » Wed May 12, 2010 6:04 am
Just a quick test tip on this one, folks. A problem like this demonstrates the importance of being able to quickly "Reverse FOIL" a quadratic equation. This is a great technique for anybody that wants to solve this without introducing new variables. The technique, in brief, is as follows.

Let's take the equation X^2 + 5X + 6 = 0 as an example.

1) Set up your factored form as ( X + __ ) ( X + __ ) = 0

2) Then simply fill in the blanks with constants that ADD UP to 5 (the X coefficient) and have a PRODUCT of 6 (the constant). In this simple example, they will be 2 and 3.

3) The roots are the OPPOSITE of each of those constants (because the equation will be true when each linear factor equals 0). So the solutions to this example are -2 and -3.

This takes bit of trial and error, but most quadratics that appear on the GMAT can be "Reverse FOILed" pretty easily. They just seem to enjoy testing us on this skill.

Note that these values CAN BE and often are negative.

(You'll notice, also, that this would become much harder if there is a coefficient greater than one on the squared term. It can be done, but is a bit trickier, so I will not go down that road here.)

Now ... for this question, you know that your factored form is going to be ( X - 3 ) ( X + __ )

Determine what your other constant will be, based on each answer choice. To do this, divide the constant by -3, and then see if that number minus 3 gives you the X coeffiecient.

A) -5 works
B) -6 works
C) non-integer, therefore can't reverse foil easily.
D) 4 works
E) 18/-3 gives you 6, but you end up with the wrong X coeffiecient. -3 must not be a solution.

Since the second root is the OPPOSITE of the constant we've come up with, and must be a negative even integer, the only answer choice that works out for us is D.

Hope that's useful!

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