Advanced Quant - Problem #125

[shailendra.sharma]

If x > 0, what is the least possible value of x + 4/x

A) 0
B) 1
C) 2
D) 3
E) 4

Reason for this post is to verify my approach for solving this question --

For each answer choice I set up the quadratic equation, say for answer choice D

x + 4/x = 3 ==> x^2 -3x + 4 = 0
I computed b^2 - 4ac to find if this equation has any real solutions. That is b^2 - 4ac >= 0

For only answer choice E, it was true. And once, I did this setup for one answer choice - checking other answer choice was too quick - answer choices control only b^2 component and 4ac was constant across all choices (which is 16).

Please comment on the approach.

Edit: Added the answer choices.

[jlucero]

First off, please post the full answer choices so we can help out more students that happen upon this thread.

Second, that's an incredibly algebraic approach to this question that I never would have thought of, but you actually could have taken it one step further. Since you're trying to find when b^2 - 4ac >= 0, you could have searched for when b^2 >= 4ac. Note that from your quadratic equations, a = 1, b = the answer (x), and c = 4. Plugging those in, x^2 > 4*1*4, x^2 >= 16, x >= 4. At the same time, I think it would be much faster to plug in numbers here:

x = 1, x + 4/x = 5
x = 2, x + 4/x = 4
x = 3, x + 4/x = 4.33
x = 4, x + 4/x = 5
x = 5, x + 4/x = 5.8
Note I can stop once I realize that x is going to get larger and larger, and the smaller fraction won't compensate for the larger x. I think this is a far easier method with much less computation than what you did.

[shailendra.sharma]

Well, my idea was not to compare two approaches - one in book and the way I did - I am not sure if my post alluded that.

I am just trying to validate whether my approach was mathematically right in these type of problem.

And yes, after identifying relation between B^2-4AC -- I could jump to choice E (correct choice) directly.

[tim]

"Mathematically right" as a term is a little ambiguous. :) Your approach worked, but it was not purely mathematical in the sense that you only got lucky because (as you say*) only one answer choice gave you real values. What if two or more answer choices had given you real values? Presumably you would have chosen the smallest one. However, that is also not a purely mathematical solution, as this method would rely on being given a set of answer choices. What you would want to do if you were really intent on a provable algebraic solution is to look for where the discriminant is 0, i.e. where there is a double root.

*Note that your solution method is particularly ironic given the fact that it relies so heavily on the specific answer choices, yet you didn't give us the answer choices even after Joe asked you to!

[shailendra.sharma]

Edited my first post to add the answer choices.

Though I am not really convinced on reasoning for undermining the solution.

In my opinion, my solution had a mathematical reasoning + number picking -- I am not trying to solve it with pure mathematics. But I selected the discriminant approach, as I could see the equation was of quadratic form and for it to have a value (whether least or highest), equation need to have some real solutions and hence I went into B^2 - 4AC path.

Please understand I am not trying to undermine the solution given in Advanced Quant book, in any ways. That book has been written by best minds and I do not have any reason to compete with those -- nothing to gain. Unfortunately, this thread has gone into unwanted direction, whereas my aim for post was to gain certain insights !

My aim is to understand, what could go wrong by choosing a discriminant approach, such as you can give me insights discriminant is not going to work for these type of equations. As you said, even if there were two solutions I could have picked the least value as required by the question.

[RonPurewal]

Because this is a multiple-choice problem with answer choices, your approach is perfectly valid: the "least possible value" is just the smallest answer choice that actually has solutions. So, yeah. Legitimate.

By the way, this problem would never appear on the gmat in this form, because it's exactly the kind of thing that the gmac people seek to avoid: namely, it's a problem that is trivially easy if you know calculus (which isn't required on the gmat), but much more annoying if you don't.
The gmac people are very, very good at writing problems on which more "advanced" math (calculus, trigonometry, etc.) not only aren't necessary, but don't even really help. So you won't see a problem like this one.

Nevertheless, the problem still has value, in the sense that it can get you to think about various approaches, such as the two disparate ones in this thread.

[shailendra.sharma]

Thanks Ron. It makes sense now.

[jnelson0612]

Great!