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Quadratic Equation

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Quadratic Equation

by usfall13ivy » Wed May 02, 2012 7:55 pm
What is the value of 6x^2 + x - 1 ?
(1) x(x + 3) = 0
(2) x = -3

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

The Source is Gmathacks.com
The official answer is B.
I think the answer is C since where you divide the first statement by x, you get x = -3 and substituting that, you get a single answer for the equation. Can someone please explain the flaw in my reasoning or assert that the answer is indeed C.
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by shantanu86 » Wed May 02, 2012 9:12 pm
Hi usfall13ivy,

The correct answer is [spoiler][/spoiler]

f(x) = 6x^2 + x -1
=>(2x+1)(3x-1)

Now from (1)

x(x+3) = 0
This is satisfied when x = 0 and x = -3
for x=0, f(x) = -1;
for x=-3, f(x) = 50
hence, no unique value.. NOT sufficient

From (2)
x = -3
=> f(x) = 50
Unique value therefore sufficient

Hence the answer is [spoiler][/spoiler]

Hope it helps!!
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by Stuart@KaplanGMAT » Wed May 02, 2012 11:02 pm
usfall13ivy wrote:I think the answer is C since where you divide the first statement by x, you get x = -3 and substituting that, you get a single answer for the equation. Can someone please explain the flaw in my reasoning or assert that the answer is indeed C.
Hi!

There are two problems with your analysis, one of which may just have been a typo.

First, the minor issue: (C) is correct if you require both statements in combination; based on the rest of your post, I'm going to assume that you meant that you thought the correct answer should be (D), either statement is good enough by itself.

The bigger issue, however, is a common mistake when simplifying equations. You divided both sides by x to simplify, but you forgot that there's one case in which we're not allowed to divide - when the quotient is 0. Since it's possible that x=0, you can't simply cancel it out of the left side.

There are two ways to avoid making the mistake:

1) when you divide both sides by a variable, break it up into two cases: x=0 and x(not=)0. For the first case you have one possible solution, for the second you'll create another possible value for x.

2) just don't divide through by x. Instead, use the product of 0 rule: the only way to get a product of 0 is to multiply by 0. Accordingly, if:

x(x+3) = 0,

then either:

x = 0
or
x + 3 = 0, i.e. x = -3.

The 2nd option is almost certainly the simpler way to go.
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