If this were something other than the GMAT, I'd use calculus to solve the question (i.e., find the derivative of x + 4/x and locate the value of x such that the slope = 0)knight247 wrote:If x>0, what is the least possible value of (x)+(4/x)?
(A)0
(B)1
(C)2
(D)3
(E)4
OA is E
But given that it's a GMAT question, we'll use some number sense.
First, if x>0, we know that 4/x>0
This means that x + (4/x) > 0
So, we can already rule out A
Now let's try a few values of x.
If x=1 then (x)+(4/x)= 5
If x=2 then (x)+(4/x)= 4
If x=3 then (x)+(4/x)= 4 1/3
If x=4 then (x)+(4/x)= 5
We don't need to check any values of x greater than x=4 since these values will make the value of (x)+(4/x) even bigger.
At this point, the question becomes, "Are there any values of x such that (x)+(4/x) is less than 4 (if x>0)?" For example, perhaps there's a value of x between 1 and 2 that yields an even smaller value of (x)+(4/x).
The answer to the question is no, but it's hard to prove unless we use some calculus or we know how the graph of y = (x)+(4/x) looks. Both of these, however, are beyond the scope of the GMAT.
At this point, I'd look at my previous results and hope that the answer is E
But to be more precise, we could check each answer choice.
D) is there a value of x such that (x)+(4/x) = 3?
To solve this, we'll multiply both sides by x to get x^2 + 4 = 3x
If we rewrite this as x^2 - 3x + 4 = 0 we can see that this equation has no solution.
To prove that this equation has no solution, we'll need to use the quadratic formula
Aside: I have never seen an official GMAT question that required the quadratic formula, so this next part may be out of scope.
The quadratic formula says that, if ax^2 + bx + c = 0, then x = [-b + sqrt(b^2 - 4ac)] / 2a
For this question, we need only focus on this part: sqrt(b^2 - 4ac)
In the equation x^2 - 3x + 4 = 0, a=1, b=-3 and c=4
So, we get: sqrt(b^2 - 4ac) = sqrt((-3)^2 - 4(1)(4))
= sqrt(-7)
Since we cannot evaluate sqrt(-7), we can conclude that x^2 - 3x + 4 = 0 has no solution.
This means we can eliminate answer choice D.
We can use similar logic to eliminate B and C as well, but I think it's safe to say that this question might be out of scope.
Nonetheless, the answer is E
Cheers,
Brent
