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Source: — Data Sufficiency |

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by lunarpower » Thu Apr 24, 2008 2:08 am
i'm not sure i understand what statements (1) and (2) are supposed to say here; please post clear and unambiguous versions of statements (1) and (2).

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in any case, here's a start:

since you're looking for an intersection point, you know that the y coordinates have to be the same. therefore, ax - b must equal x^2 + b, which means that ax - x^2 = 2b, or, x(a - x) = 2b.

can't get any farther without the statements.
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by Stuart@KaplanGMAT » Thu Apr 24, 2008 11:56 am
Whenever you post inequalities, make sure you click the "disable HMTL in this post" box below the text window. HTML uses < and > for tags and if you have HTML on your inequalities will get all squishy!

Similarly, if you ever see a smiley in one of your posts and didn't mean to put one there, it's probably because you typed 8) and didn't disable Smilies.
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by mayank_gupta123 » Tue Oct 14, 2008 12:41 pm
Interesting question!

If there is at all an intersection, it would be a single point (obvious)

In that case

=> aX - b = X2 + b

=> X2 – aX + 2b = 0


For a real solution of X,

i.è. B2 – 4AC > 0



i.e. (-a)2 – 4 * 2b >0

i.e. b < a2/8



Irrespective of any sign of a, if b is less than zero, that would do the needful, therefore B should be the right answer.


Regards,
Mayank
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by cubicle_bound_misfit » Tue Oct 14, 2008 9:33 pm
mayank,

let go parabola but whenever a single degree equation cuts an n degree equation there will be a n point cut.
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by 69aero69 » Fri Oct 17, 2008 5:44 am
since I cannot see the options, I don't know the final response but here is a way to solve it:

x^2 + B = Ax - B
<=> x^2 - Ax + 2B = 0 <=>
<=> x=[A +- sqrt(A^2 - 8B)]/2

therefore we have to analyse de root:
if A^2 - 8B < 0 - No intersection
if A^2 - 8B = 0 - One intersection
if A^2 - 8B > 0 - two intersections

To anwser the question we need to know if there is at least one inters, i.e., A^2 - 8B >=0,

Thus,
[A >= sqrt(8B) or A < -sqrt(8B) ] and b>0

Now we can only anwser depending on the info given..
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