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

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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, 20:24
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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
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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, 01:02
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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 13 Oct 2018, 05:51
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
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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 21 Oct 2018, 02:10
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 &nbs [#permalink] 21 Oct 2018, 02:10
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