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# If k is an integer and x(x – k) = k + 1, what is the value of x?

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SVP
Joined: 06 Nov 2014
Posts: 1877
Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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18 Jul 2016, 20:41
1
Bunuel 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

Kudos for a correct solution.

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
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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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18 Nov 2016, 08:48
Bunuel 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

Kudos for a correct solution.

I picked A...
x<k
so we have few options:
1)
x negative
k negative
x-k = negative -> x*(x-k) = positive.
since k+1 must be positive, k must be either zero, or a non-integer. but since we assumed k is negative, then it doesn't work...

2)
x positive
k positive
x-k - negative
negative * positive = negative
it means that k+1 must be negative, meaning that k must be < -1. but we assumed k is positive. doesn't work...

3) we can't assume x positive and k negative, as we are told that x<k

4.
x negative
k positive

x-k - negative
multiply by x (negative) = positive
k+1 is positive.

let's test some values:
x=-1
k=1
-1(-1-1) = -1*-2 = 2 = 1+1 - works

x=-2
k=1
-2(-2-1) = 6
but k=1. so doesn't work

we can test more options, but it will not work unless x=-1. A is sufficient.

2. x=3-k
x(x – k) = k + 1
k=3-x

x(x-3+x) = 3-x+1
x(2x-3) = 4-x
2x^2 -3x = 4-x
2x^2 -2x -4 = 0 | divide by 2
x^2 -x -2 = 0
(x+1)(x-2)=0
x=2; x=-1

1st option
x=2
x = 3 – k
2=3-k
k=1
x(x – k) = k + 1 = 2(2-1) = 1+1
2=2

2nd option:
x=-1
x = 3 – k
-1 = 3-k
k=4
x(x – k) = k + 1 = 2(2-1) = 1+1
-1(-1-4) = 4+1 = 5=5
works
2 outcomes for x - not sufficient.

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If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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18 Nov 2016, 09:11
Bunuel 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

Kudos for a correct solution.

From statement 1>
x(x-k) = k+1 => x^2-xk-k-1 = 0
=> x = (k+(k^2+4k+2^2)^1/2)/2 or x = (k- (k^2+4k+2^2)^1/2)/2
Since, k^2+4k+2^2 = (k+2)^2
By solving this we get
either x= k+1 or x= -1; but x <k, therefore x can't be k+1
Hence, x = -1 => Sufficient

From statement 2 >
we can solve quadratic for k and get values 2 values for x and k each. Therefore it is insufficient

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Joined: 23 Jan 2013
Posts: 559
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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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18 Nov 2016, 23:42
(x-1)*(x+1)-k(x+1)=0

(x+1)*(x-1-k)=0

x+1=0, x=-1
OR
x-1-k=0, x=k+1

St.1 says that second scenario is impossible, so x=-1

A
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If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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31 Mar 2017, 23:27
Bunuel 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

Kudos for a correct solution.

statement 1:
x(x-k)=k+1
x^2-xk=k+1
x^2-1=k+xk
(x-1)(x+1)-(x+1)(k)=0
(x+1)[x-1-k]=0

this means x=-1; or x=k+1 not valid since x<k(given)
in sufficient as we cannot say anything about k when x=-1

statement 2:
putting x=3-k given 2 values x=-1;k=4
x=2;k=1
insufficient

taking both together:
x=-1;k=4
solution

give kudos if you liked this explanation
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If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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31 Mar 2017, 23:35
Engr2012 wrote:
Swaroopdev wrote:
Hi Bunuel,

None of the above solutions show A as the answer, is the OA wrong ? If not, could you please explain the solution ?

Thanks.

I think everyone is missing out on a very important information.

Lets analyse the question:

Given equation: x(x-k) = k+1 ----> $$x^2-kx-k-1=0$$ --->$$x = \frac {k\pm \sqrt{k^2-4(-k-1)}}{2}$$

Thus you get 2 values of x as =$$\frac {k\pm(k+2)}{2}$$ ---> x = k+1 or x=-1

Per statement 1, x<k ---> x $$\neq$$k+1 ---> x = -1 . Thus this statement IS SUFFICIENT

Per statement 2, x = 3-k ---> susbtituting in the above given equation, you get, k = 4 or 1 ---> x = -1 or 2. Thus this statement is NOT sufficient.

This question is a classic "C-Trap" question.

how can we conclude through statement A when x=-1 x<k
say x=-1 and k=-2 then also they satisfy the equation and
when x=-1 k=2 they also satisfy the equation

then how can we find the values where k>x always????????

thus answer is C when x=-1 k also takes unique value k=4
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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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31 Mar 2017, 23:38
OptimusPrepJanielle wrote:
Bunuel 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

Kudos for a correct solution.

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

how can we conclude through statement A when x=-1 x<k
say x=-1 and k=-2 then also they satisfy the equation and
when x=-1 k=2 they also satisfy the equation

then how can we find the values where k>x always????????

thus answer is C when x=-1 k also takes unique value k=4
Intern
Joined: 05 Dec 2016
Posts: 8
If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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12 Apr 2017, 00:15
Bunuel, I was wondering about Statement 2. Using x = 3-k, we could calculate as follows:

(3-k) (3-k-k) = k+1 ==> k^2 - 5k + 4 = 0 ; k= 4, 1

Using K=4 in the original equation x(x-k) = k+1; we get:

x^2 - 4x - 5 =0 ==> x= 5 , -1

Using K=1 in the original equation x(x-k) = k+1; we get:

x^2 - x - 2 = 0 ==> x = 2 , -1

Since -1 is the only solution of x that satisfies both values of k in the original equation, can we not claim that statement 2 is sufficient?
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Joined: 02 Sep 2009
Posts: 52296
Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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12 Apr 2017, 01:13
dg88074 wrote:
Bunuel, I was wondering about Statement 2. Using x = 3-k, we could calculate as follows:

(3-k) (3-k-k) = k+1 ==> k^2 - 5k + 4 = 0 ; k= 4, 1

Using K=4 in the original equation x(x-k) = k+1; we get:

x^2 - 4x - 5 =0 ==> x= 5 , -1

Using K=1 in the original equation x(x-k) = k+1; we get:

x^2 - x - 2 = 0 ==> x = 2 , -1

Since -1 is the only solution of x that satisfies both values of k in the original equation, can we not claim that statement 2 is sufficient?

We have: x(x – k) = k + 1 and x = 3 – k.
k = 1 and x = 2 OR k = 4 and x = -1. Both sets satisfy the given inequalities.
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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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12 Apr 2017, 02:13
Bunuel wrote:
dg88074 wrote:
Bunuel, I was wondering about Statement 2. Using x = 3-k, we could calculate as follows:

(3-k) (3-k-k) = k+1 ==> k^2 - 5k + 4 = 0 ; k= 4, 1

Using K=4 in the original equation x(x-k) = k+1; we get:

x^2 - 4x - 5 =0 ==> x= 5 , -1

Using K=1 in the original equation x(x-k) = k+1; we get:

x^2 - x - 2 = 0 ==> x = 2 , -1

Since -1 is the only solution of x that satisfies both values of k in the original equation, can we not claim that statement 2 is sufficient?

We have: x(x – k) = k + 1 and x = 3 – k.
k = 1 and x = 2 OR k = 4 and x = -1. Both sets satisfy the given inequalities.

Oh it makes sense now. Thanks a lot Bunuel
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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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21 Apr 2017, 15:18
Bunuel 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

Kudos for a correct solution.

st1: x^2-xk=k+1
(x-1)(x+1)=K(x+1)
x+1= 0
x=-1, k=0
x=k+1( irrelevant since we know k>x)

st2: x=3-k
x=2, k=1 or x=-1, k=4 (insufficient on its own)
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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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07 Jul 2017, 05:34
smanujahrc wrote:
OptimusPrepJanielle wrote:
Bunuel 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

Kudos for a correct solution.

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

how can we conclude through statement A when x=-1 x<k
say x=-1 and k=-2 then also they satisfy the equation and
when x=-1 k=2 they also satisfy the equation

then how can we find the values where k>x always????????

thus answer is C when x=-1 k also takes unique value k=4

You are given that x=-1 OR x=1+k,

Statement 1 says that x<k, so how can x=1+k be true? x=1+k will ALWAYS make x>k, thus going against the information mentioned in statement 1. Thus, you MUST reject x=1+k, leaving you with x=-1 as the only solution from statement 1. Hence this statement is sufficient.
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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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16 Dec 2017, 05:20
Finally some sane question.
x(x-k) = k+1
=> x^2 - xk = k+1
=> X^2-1 - xk - k = 0
=> (x-1)(x+1) - k(x+1) = 0
=> (x+1((x-1-k)=0
POssible values x=-1, or, x=k+1

(1) x<k. Therefore, the only value is x=-1. Sufficient
(2) x=3-k.
Checking two possible values of x, x= k+1 => 3-k =k+1 => k=1 & x=2.
if x=-1, => 3-k=-1, => k=4, x=-1.
Both values of x are possible. Ambiguous. Insufficient

Hence A
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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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10 Apr 2018, 12:38
Hello, everybody!

I think one of quickest ways to solve it is by using a parabola method and Vieta's formulas ( x1*x2 = c/a; x1 + x2 = -b/a).

Case 1:

1) So, we found x1*x2 = -k-1; and x1 + x2 = k.

Meaning, x1 = k +1,
x2 = -1.

2) Our parabola has arms up and two possible x-intercepts: x = -k-1; and x = -1.

If (-k-1) lays to the right of (-1), then -k-1 > -1.
So, k <0. And x<k.

Then, only possible solution is x2 = -1 (because whenever you add 1 to a negative number - it will become bigger, not smaller than it was; meaning x1 does not match the case). Sufficient.

If (-k-1) lays to the left of (-1), then -k-1 < -1.
So,k >0. And x < k.

Then, the only possible solution is x2 = -1 (becuse whenever you add 1 to a positive number - it will become bigger than it was; meaning x1 does not match the case). Sufficient.

Case 2:
x = 3-k.
After soving the equation: k = 4; k = 1. So, 2 choices for x. Not sufficient.
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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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07 Nov 2018, 09:16
Hi guys,

Im pretty much new to this forum and I'm trying to practice DS questions. Can someone please tell me how to know which one is the correct answer. As I'm able to see pretty much everyone coming up with their own version of the answer. Please help. Thanks.
Re: If k is an integer and x(x – k) = k + 1, what is the value of x? &nbs [#permalink] 07 Nov 2018, 09:16

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