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

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Intern
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For which value of k does the following pair of equations yield a [#permalink]

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New post 10 Nov 2015, 21:24
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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 - 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
[Reveal] Spoiler: OA

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Re: For which value of k does the following pair of equations yield a [#permalink]

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New post 23 Jun 2016, 02:02
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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

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Re: For which value of k does the following pair of equations yield a [#permalink]

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New post 31 Aug 2017, 22:44
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Re: For which value of k does the following pair of equations yield a   [#permalink] 31 Aug 2017, 22:44
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