Mo2men wrote:If k is an integer and x(x - k) = k + 1, what is the value of x?

(1) x < k
(2) x = 3 - k

Given: x(x-k) = k + 1
x^2 - xk -k - 1 = 0
x^2 - 1 -k(x+1) = 0
(x-1)(x+1) - k (x+1) = 0
(x+1)(x-1-k) = 0 - (i)

Hence x = - 1 or x = 1+k

Required: x = ?

Statement 1: x < k
From this, we can say that ≠ k+1
Hence x can only take the value -1
SUFFICIENT

Statement 2: x = 3-k
On substituting the value of k in the solutions for the equation, we get
3-k = - 1 and 3-k = 1+k
Hence k = 4 and k = 1
INSUFFICIENT

Correct Option: A

A very good question which can trap you in choosing the option C

OptimusPrep wrote:

Mo2men wrote:If k is an integer and x(x - k) = k + 1, what is the value of x?

(1) x < k
(2) x = 3 - k

Given: x(x-k) = k + 1
x^2 - xk -k - 1 = 0
x^2 - 1 -k(x+1) = 0
(x-1)(x+1) - k (x+1) = 0
(x+1)(x-1-k) = 0 - (i)

Hence x = - 1 or x = 1+k

Required: x = ?

Statement 1: x < k
From this, we can say that ≠ k+1
Hence x can only take the value -1
SUFFICIENT

Statement 2: x = 3-k
On substituting the value of k in the solutions for the equation, we get
3-k = - 1 and 3-k = 1+k
Hence k = 4 and k = 1
INSUFFICIENT

Correct Option: A

A very good question which can trap you in choosing the option C

Thanks Ankur for you help.

I have a question regarding statement 1. I do not understand how it satisfies the condition that x<k when x=-1. We do not know what the value of k except that we substitute x=-1 in equation and get k=0 then K>x.

Am I right?

Mo2men wrote:If k is an integer and x(x - k) = k + 1, what is the value of x?

(1) x < k
(2) x = 3 - k

Statement 1: x<k
Case 1: k=0
Substituting k=0 into x(x - k) = k + 1, we get:
x(x - 0) = 0 + 1
x² = 1
x = ±1.
Since statement 1 requires that x<k, the only viable option is x=-1.

Case 2: k=10
Substituting k=10 into x(x - k) = k + 1, we get:
x(x - 10) = 10 + 1
x² - 10x = 11
x² - 10x - 11 = 0
(x-11)(x+1) = 0
x = 11 or x=-1.
Since statement 1 requires that x<k, the only viable option is x=-1.

Case 3: k=-1
Substituting k=-1 into x(x - k) = k + 1, we get:
x(x + 1) = -1 + 1
x² + x = 0
x = 0.
Since statement 1 requires that x<k, this case is not viable.

Case 4: k=-10
Substituting k=-10 into x(x - k) = k + 1, we get:
x(x + 10) = -10 + 1
x² + 10x = -9
x² + 10x + 9 = 0
(x+9)(x+1) = 0
x = -9 or x=-1.
Since neither option for x is such that x<k, this case is not viable.

In every viable case, x=-1.
SUFFICIENT.

Statement 2: x=3-k
Substituting k = 3-x into x(x - k) = k + 1, we get:
x(x - (3-x)) = 3-x + 1
x(2x - 3) = 4 - x
2x² - 3x = 4 - x
2x² - 2x - 4 = 0
x² - x - 2 = 0
(x-2)(x+1) = 0
x=2 or x=-1.
Since x can be different values, INSUFFICIENT.

The correct answer is A.

Brent@GMATPrepNow wrote:

Mo2men wrote:If k is an integer and x(x - k) = k + 1, what is the value of x?

(1) x < k
(2) x = 3 - k

This is a challenge question I created for BTG a while back.
Here's a video solution.

Cheers,
Brent

Hi Brent,

There is not video attached.

Thanks

Mo2men wrote:I have a question regarding statement 1. I do not understand how it satisfies the condition that x<k when x=-1. We do not know what the value of k except that we substitute x=-1 in equation and get k=0 then K>x.

Am I right?

If we have

x = k + 1 or x = -1

we know that the first of these (x = k + 1) gives us

x = k + 1

k + 1 > k

so

x > k

S1 tells us that x is NOT greater than k, so x = k + 1 must NOT be the root that we're looking for. That means we want the other root: x = -1.