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Confusing problem

This topic has 2 expert replies and 4 member replies

Confusing problem

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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

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GMAT/MBA Expert

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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
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)
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

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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
I received a PM asking me to comment.

Another approach:

Let y = the least possible value.

x + 4/x = y.
x² + 4 = yx
x² - yx + 4 = 0.

The discriminant of a quadratic with equation ax²+bx+c=0 is b²-4ac.
For a quadratic to have one or more real solutions, b²-4ac ≥ 0.
In the quadratic above, a=1, b=(-y), and c=4.
Thus:
(-y)² - 4(1)(4) ≳ 0.
y² ≳ 16.

Of the answer choices, only E offers a viable value for y:
4² ≳ 16
16 ≳ 16.

The correct answer is E.

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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
(x^2+4)/x >=1 by supplying choice B) as 0 isn't possible here (all values are +ve)
since x>0, x^2+4>=x and x^2-x+4>=0
x(1,2)=[1+-Sqrt(1-16)]/2 -> Reject as Discriminant is negative.
Supply choice E) to have non-negative Discriminant

Correct answer e

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Calculus rocks! I know that we're not expected to know how to find a derivative, but it can really help!


Brent@GMATPrepNow wrote:
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
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)
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

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I have a very basic way to solve this.
0 is straight away out as explained by Brent above.
now to find least possible value of x + 4/x lets equate it to each answer choice.
So, x + 4/x = 1 ; x^2 + 4 = x ; x^2 - x + 4 =0 _____ (1)
similarly, equation with 2, 3 and 4 you get,
x^2 -2x + 4 = 0_____(2) ,
x^2 -3x + 4 = 0_____(3) ,
x^2 -4x + 4 = 0_____(4) .

As x^2 & 4 is common in all the 4 equations above the deciding factor is the value of the middle term .
As x>0 its common sense that equation 4 will yield the least value.

Hence, E.

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Thank you all the guyzzz for a wonderful explanation...

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