If A and C are positive constants, then which of the

Question

If A and C are positive constants, then which of the following equations must have at least one negative solution?

I. X^2+AX+C=0
II. X^2-AX+C=0
III. X^2+AX-C=0

A) I Only
B)II Only
C) III Only
D) II and III Only
E) I, II, and III Only
F) I and II Only
G) I have no clue what the question is asking.

HIMALAYA · Answer

E) I, II and III Only

I has both -ve solutions.
II has one -ve solution.
III has one -ve solution.

vikramm · Answer

D) II & III only.

HIMALAYA · Answer

let me change to (A) I Only but i think it should be I and III.

I has both -ve solutions.
II cannot have -ve solution.
III can have one -ve solution.

Titleist · Answer

Himalaya,

Clue for this problem. Think of the quadratic formula. Also think of what the rules are when you have a negative output vs. positive output vs. a output of 0. Hope this helps.

Titleist · Answer

Good Job Pete oA is c

HIMALAYA · Answer

how this is possible, if in (I) X^2+AX+C=0. here x must be -ve otherwise X^2+AX+C doesnot become zero.

Any OE?

vikramm · Answer

Doesn't II give 2 negative results?? Question asks for "atleast one negative solution" ... explain please.

HIMALAYA · Answer

II doesnot give any -ve solution but I gives. :roll:

suppose A=5 and C= 6 from (I) x^2+Ax+C = 0 x^2+5x+6=0 x^2+3x+2x+6=0 (x+2)(x+3)=0 x = -2, -3 from (II) x^2-Ax+C = 0 x^2-5x+6=0 x^2-3x-2x+6=0 (x-2)(x-3)=0 x = 2,3 from (III) x^2+Ax-C = 0 x^2+5x-6=0 x^2+6x-x+6=0 (x+6)(x-1)=0 x = -6, 1 so I and III give at least one -ve solution.....

vikramm · Answer

I did that twice in one day!!! :wall

petefroml · Answer

This equation doesn't necessarily have a solution. It's the case if we choose A=1 and C=1. Same for II.

Titleist · Answer

Himalaya,

You don't know if I gives one solution - because you don't know the value of the constants of A and C. You can't just pick numbers on this one and assume it will work for all cases. You must use the quadratic formula 

-b+or-Sqrt(b^2-4ac)/2a


focus on sqrt(b^2-4ac) if this output is positive then you have 2 solutions (one positive and one negative) if this equals 0 then you have only one solution - may be positive or negative. We also don't know if this calculation equals a negative number; then we would have an undefined solution since the square root of a negative number is undefined (in the case of the gmat). 

so compare the answer choices:

I) sqrt(A^C-4(1)(+C)) we don't know because A^2 could be equal to -4*1*C in which the output is zero - thus having only 1 x-intercept (neg or positive). You can make the same argument that A^2 could be less than -4*1*C in which case you would have an undefined or complex solution (no x-intercept) - we don't know here. 

II) sqrt(A^2-4*1*C) A as a neg number doesn't give us anything definitive in terms of a neg, pos, or undefined solution since you're just squaring the b coefficient

III) sqrt(A^2-4*1*(-C)) you will always derive a positive solution here because A^2-4*1*(-C) equals (A^2+4*1*C); also, you're guaranteed one neg solution here because the C coefficient is negative. 


Hope this makes sense

HIMALAYA · Answer

I will take you guys' posts later. now i am too tired. today i spent almost 5 hours in the forum and tried to solve many problems. i am taking rest for a second and going out for refereshment.........

anyway, thanks guys...

HongHu · Answer

X^2+AX+C=0
(x+A/2)^2-A/2+C=0
x=+/-sqrt(A/2-C)-A/2

X^2-AX+C=0
(x-A/2)^2-A/2+C=0
x=+/-sqrt(A/2-C)+A/2

X^2+AX-C=0
(x+A/2)^2-A/2-C=0
x=+/-sqrt(A/2+C)-A/2

Ask for "must have at least one negative solution".
For (I), if A/2>=C then it must have at least one negative solution, however if A/2<C then there's no solution.
For (II), same as I, only it may or may not have a negative solution, even if solutions exist.
For (III) since A/2+C is always greater than zero, therefore solutions exist. And one of the solutions must be negative.

If A and C are positive constants, then which of the

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