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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # For which value of k does the following pair of equations yield a uniq

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Math Expert V
Joined: 02 Sep 2009
Posts: 61385
For which value of k does the following pair of equations yield a uniq  [#permalink]

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12 00:00

Difficulty:   95% (hard)

Question Stats: 24% (02:40) correct 76% (02:42) wrong based on 46 sessions

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For which value of k does the following pair of equations yield a unique solution for x, such that the solution is positive?

$$(x - k)^2 - y^2 = 8$$
$$(x + 2k)^2 + (y^2 - k^2) = 0$$

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

(B) $$\frac{\sqrt{7}}{4}$$

(C) $$\sqrt{2}$$

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

(E) None of these

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VP  V
Joined: 19 Oct 2018
Posts: 1301
Location: India
Re: For which value of k does the following pair of equations yield a uniq  [#permalink]

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

$$y^2= (x-k)^2-8$$......(1)

$$(x + 2k)^2 + (y^2 - k^2) = 0$$

From (1)

$$(x + 2k)^2 + [(x-k)^2-8 - k^2] = 0$$

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

We have a unique solution of x; hence, b^2-4ac=0

k^2 - 4 ( 2k^2 - 4 )= 0

k = $$\frac{+4}{\sqrt{7}}$$ or $$\frac{-4}{\sqrt{7}}$$

Also, solution is positive; hence, their sum is also positive. k must be negative

Only possible value of k is $$\frac{-4}{\sqrt{7}}$$

Bunuel wrote:
For which value of k does the following pair of equations yield a unique solution for x, such that the solution is positive?

$$(x - k)^2 - y^2 = 8$$
$$(x + 2k)^2 + (y^2 - k^2) = 0$$

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

(B) $$\frac{\sqrt{7}}{4}$$

(C) $$\sqrt{2}$$

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

(E) None of these

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VP  P
Joined: 24 Nov 2016
Posts: 1224
Location: United States
Re: For which value of k does the following pair of equations yield a uniq  [#permalink]

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Bunuel wrote:
For which value of k does the following pair of equations yield a unique solution for x, such that the solution is positive?

$$(x - k)^2 - y^2 = 8$$
$$(x + 2k)^2 + (y^2 - k^2) = 0$$

(A) $$\frac{-4}{\sqrt{7}}$$
(B) $$\frac{\sqrt{7}}{4}$$
(C) $$\sqrt{2}$$
(D) $$\frac{4}{\sqrt{7}}$$
(E) None of these

x = unique positive solution, so $$b^2-4ac=0$$ and b<0;

$$(x - k)^2 - y^2 = 8$$
$$(x + 2k)^2 + (y^2 - k^2) = 0$$
$$[x^2+k^2-2xk-y^2=8]+[(x + 2k)^2 + y^2 - k^2 = 0]$$
$$[x^2-2xk]+[(x + 2k)^2]=8…x^2-2xk+x^2+4k^2+4xk=8$$
$$2x^2+2xk+4k^2=8…x^2+xk+2k^2=4…x^2+xk+(2k^2-4)=0$$
$$b^2-4ac=0…(xk)^2-4(x^2)(2k^2-4)=0…(xk)^2=4(x^2)(2k^2-4)$$
$$k^2=4(2k^2-4)…8k^2-16-k^2=0…7k^2=16…k^2=16/7$$
$$|k|=[4/√7,-4/√7]…[x^2+xk+(2k^2-4)=0]…[xk]=[b<0]…k=-4/√7$$

Ans (A)
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For which value of k does the following pair of equations yield a uniq  [#permalink]

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chetan2u

Can you help me with this question please.
Intern  B
Joined: 18 Feb 2019
Posts: 16
Re: For which value of k does the following pair of equations yield a uniq  [#permalink]

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exc4libur wrote:
Bunuel wrote:
For which value of k does the following pair of equations yield a unique solution for x, such that the solution is positive?

$$(x - k)^2 - y^2 = 8$$
$$(x + 2k)^2 + (y^2 - k^2) = 0$$

(A) $$\frac{-4}{\sqrt{7}}$$
(B) $$\frac{\sqrt{7}}{4}$$
(C) $$\sqrt{2}$$
(D) $$\frac{4}{\sqrt{7}}$$
(E) None of these

x = unique positive solution, so $$b^2-4ac=0$$ and b<0;

$$(x - k)^2 - y^2 = 8$$
$$(x + 2k)^2 + (y^2 - k^2) = 0$$
$$[x^2+k^2-2xk-y^2=8]+[(x + 2k)^2 + y^2 - k^2 = 0]$$
$$[x^2-2xk]+[(x + 2k)^2]=8…x^2-2xk+x^2+4k^2+4xk=8$$
$$2x^2+2xk+4k^2=8…x^2+xk+2k^2=4…x^2+xk+(2k^2-4)=0$$
$$b^2-4ac=0…(xk)^2-4(x^2)(2k^2-4)=0…(xk)^2=4(x^2)(2k^2-4)$$
$$k^2=4(2k^2-4)…8k^2-16-k^2=0…7k^2=16…k^2=16/7$$
$$|k|=[4/√7,-4/√7]…[x^2+xk+(2k^2-4)=0]…[xk]=[b<0]…k=-4/√7$$

Ans (A)

Hi ,
[x = unique positive solution, so $$b^2-4ac=0$$ and b<0 ]

Can you explain me why b should be < 0?
VP  V
Joined: 19 Oct 2018
Posts: 1301
Location: India
Re: For which value of k does the following pair of equations yield a uniq  [#permalink]

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-b is the sum of roots of the equation. It's given in the question that solutions or roots of the equation are positive.

Hence, -b>0 or b<0

Ahmed9955 wrote:
exc4libur wrote:
Bunuel wrote:
For which value of k does the following pair of equations yield a unique solution for x, such that the solution is positive?

$$(x - k)^2 - y^2 = 8$$
$$(x + 2k)^2 + (y^2 - k^2) = 0$$

(A) $$\frac{-4}{\sqrt{7}}$$
(B) $$\frac{\sqrt{7}}{4}$$
(C) $$\sqrt{2}$$
(D) $$\frac{4}{\sqrt{7}}$$
(E) None of these

x = unique positive solution, so $$b^2-4ac=0$$ and b<0;

$$(x - k)^2 - y^2 = 8$$
$$(x + 2k)^2 + (y^2 - k^2) = 0$$
$$[x^2+k^2-2xk-y^2=8]+[(x + 2k)^2 + y^2 - k^2 = 0]$$
$$[x^2-2xk]+[(x + 2k)^2]=8…x^2-2xk+x^2+4k^2+4xk=8$$
$$2x^2+2xk+4k^2=8…x^2+xk+2k^2=4…x^2+xk+(2k^2-4)=0$$
$$b^2-4ac=0…(xk)^2-4(x^2)(2k^2-4)=0…(xk)^2=4(x^2)(2k^2-4)$$
$$k^2=4(2k^2-4)…8k^2-16-k^2=0…7k^2=16…k^2=16/7$$
$$|k|=[4/√7,-4/√7]…[x^2+xk+(2k^2-4)=0]…[xk]=[b<0]…k=-4/√7$$

Ans (A)

Hi ,
[x = unique positive solution, so $$b^2-4ac=0$$ and b<0 ]

Can you explain me why b should be < 0? Re: For which value of k does the following pair of equations yield a uniq   [#permalink] 22 Dec 2019, 02:26
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