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

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New post 18 Jul 2016, 21:41
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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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New post 18 Nov 2016, 09:48
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.

answer is 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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New post 18 Nov 2016, 10:11
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


Therefore, answer is 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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New post 19 Nov 2016, 00:42
(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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New post 01 Apr 2017, 00:27
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.


correct answer is C

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

therefore answer should be C


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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New post 01 Apr 2017, 00:35
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.

A is the correct answer.

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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New post 01 Apr 2017, 00:38
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
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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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New post 12 Apr 2017, 01:15
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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Re: If k is an integer and x(x – k) = k + 1, what is the value of x?  [#permalink]

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New post 12 Apr 2017, 02:13
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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New post 12 Apr 2017, 03:13
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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New post 21 Apr 2017, 16:18
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)
answer is 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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New post 07 Jul 2017, 06:34
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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New post 16 Dec 2017, 06:20
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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New post 10 Apr 2018, 13:38
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.
Re: If k is an integer and x(x – k) = k + 1, what is the value of x? &nbs [#permalink] 10 Apr 2018, 13:38

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