GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 15 Jun 2019, 22:32

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

For which value of k does the following pair of equations yield a

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Intern
Intern
avatar
Joined: 03 Jun 2013
Posts: 8
For which value of k does the following pair of equations yield a  [#permalink]

Show Tags

New post 10 Nov 2015, 21:24
9
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

30% (02:41) correct 70% (03:04) wrong based on 81 sessions

HideShow timer Statistics

For which value of k does the following pair of equations yield a unique solution for x, such that the solution is positive?
(x - k)2 - y2 = 8
(x + 2k)2 + (y2 - k2) = 0

A) \(\frac{4}{\sqrt{7}}\)

B) root7/ 4

C) root 2

D)\(\frac{-4}{\sqrt{7}}\)

E) None of these
Bunuel can you help
Manager
Manager
avatar
Joined: 20 Aug 2011
Posts: 126
Re: For which value of k does the following pair of equations yield a  [#permalink]

Show Tags

New post 23 Jun 2016, 02:02
3
Equation 1: (x - k)2 - y2 = 8
Equation 2: (x + 2k)2 + (y2 - k2) = 0

Simplifying Equation 1: x^2 - 2kx + k^2 -y^2 = 8
Simplifying Equation 2: x^2 + 4kx + 4k^2 + y^2 - k^2=0

Adding equations 1 & 2, we get

2x^2 + 2kx +4k^2 = 8
2x^2 + 2kx +4k^2 - 8 = 0

x^2+ kx + 2k^2 - 4 = 0

We can use the concepts of quadratic equations in this question. For a solution to be unique, the discriminant (b^2 - 4ac) should be 0.
in this case a = 1, b = k, & c = 2k^2 - 4
So, k^2 - 4(2k^2 - 4) = 0
or, 7k^2 = 16
so k = +4/✓7 or -4/✓7
Another concept of quadratic equation is that x = (-b + ✓discriminant)/2a or x = (-b - ✓discriminant)/2a
in this case a = 1, b = k, and discriminant is zero
so we have x is - k/2
Since x is unique and positive, k should be negative, which gives us k = -4/ ✓7

So, option D
Manager
Manager
avatar
B
Joined: 18 Feb 2018
Posts: 112
Re: For which value of k does the following pair of equations yield a  [#permalink]

Show Tags

New post 13 Oct 2018, 06:51
blink005 wrote:
Equation 1: (x - k)2 - y2 = 8
Equation 2: (x + 2k)2 + (y2 - k2) = 0

Simplifying Equation 1: x^2 - 2kx + k^2 -y^2 = 8
Simplifying Equation 2: x^2 + 4kx + 4k^2 + y^2 - k^2=0

Adding equations 1 & 2, we get

2x^2 + 2kx +4k^2 = 8
2x^2 + 2kx +4k^2 - 8 = 0

x^2+ kx + 2k^2 - 4 = 0

We can use the concepts of quadratic equations in this question. For a solution to be unique, the discriminant (b^2 - 4ac) should be 0.
in this case a = 1, b = k, & c = 2k^2 - 4
So, k^2 - 4(2k^2 - 4) = 0
or, 7k^2 = 16
so k = +4/✓7 or -4/✓7
Another concept of quadratic equation is that x = (-b + ✓discriminant)/2a or x = (-b - ✓discriminant)/2a
in this case a = 1, b = k, and discriminant is zero
so we have x is - k/2
Since x is unique and positive, k should be negative, which gives us k = -4/ ✓7

So, option D


Can you please explain why x=-k/2 and why k should be -ve
Manager
Manager
avatar
D
Joined: 17 May 2015
Posts: 248
Re: For which value of k does the following pair of equations yield a  [#permalink]

Show Tags

New post 21 Oct 2018, 03:10
GittinGud wrote:
blink005 wrote:
Equation 1: (x - k)2 - y2 = 8
Equation 2: (x + 2k)2 + (y2 - k2) = 0

Simplifying Equation 1: x^2 - 2kx + k^2 -y^2 = 8
Simplifying Equation 2: x^2 + 4kx + 4k^2 + y^2 - k^2=0

Adding equations 1 & 2, we get

2x^2 + 2kx +4k^2 = 8
2x^2 + 2kx +4k^2 - 8 = 0

x^2+ kx + 2k^2 - 4 = 0

We can use the concepts of quadratic equations in this question. For a solution to be unique, the discriminant (b^2 - 4ac) should be 0.
in this case a = 1, b = k, & c = 2k^2 - 4
So, k^2 - 4(2k^2 - 4) = 0
or, 7k^2 = 16
so k = +4/✓7 or -4/✓7
Another concept of quadratic equation is that x = (-b + ✓discriminant)/2a or x = (-b - ✓discriminant)/2a
in this case a = 1, b = k, and discriminant is zero
so we have x is - k/2
Since x is unique and positive, k should be negative, which gives us k = -4/ ✓7

So, option D


Can you please explain why x=-k/2 and why k should be -ve


Hi GittinGud,

Equation 1 we can rearrange and write as follows:

\(y^2 = (x-k)^2 -8\)

Now, substitute the value of \(y^2\) in the equation(2).

\((x+2k)^2 + (x-k)^2 - 8 - k^2 = 0\)

After simplification, we get the following equation:

\(x^2 + xk + 2k^2 - 4 = 0\)

In order to have a unique solution of x, the discriminant(D) should be equal to 0. We know for a standard quadratic equation \(D = b^2 - 4ac.\)

Using the above formula we have

\(k^2 - 4*1*(2k^2 - 4) = 0 \Rightarrow k^2 - 8k^2 + 16 = 0 \Rightarrow 7k^2 = 16\)
\(k = \pm\frac{4}{\sqrt{7}}\)

One more condition is provided in the question. The value of x has to be positive.

Note that for any given quadratic equation \(ax^2 + bx +c = 0\) has two roots and given by following equation:

Roots = \(\frac{ -b \pm \sqrt{D}}{2a}\) .

If you refer the quadratic equation in x is as follows:

\(x^2 + xk + 2k^2 - 4 = 0\)

We have \(a= 1, b = k, c = 2k^2 - 4, D = 0\)

By using the above formula we have:
\(x = \frac{-k \pm \sqrt{D}}{2*1} \Rightarrow = \frac{-k}{2}\)

In order to x be +ve 'k' must be -ve. Hence, \(k = \frac{-4}{\sqrt{7}}\) .

Hope this helps.

Thanks.
GMAT Club Bot
Re: For which value of k does the following pair of equations yield a   [#permalink] 21 Oct 2018, 03:10
Display posts from previous: Sort by

For which value of k does the following pair of equations yield a

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne