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

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

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10 Nov 2015, 21:24
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95% (hard)

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30% (02:41) correct 70% (03:04) wrong based on 81 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 - 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
Joined: 20 Aug 2011
Posts: 126
Re: For which value of k does the following pair of equations yield a  [#permalink]

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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
Joined: 18 Feb 2018
Posts: 112
Re: For which value of k does the following pair of equations yield a  [#permalink]

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13 Oct 2018, 06:51
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
Joined: 17 May 2015
Posts: 248
Re: For which value of k does the following pair of equations yield a  [#permalink]

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21 Oct 2018, 03:10
GittinGud 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.
Re: For which value of k does the following pair of equations yield a   [#permalink] 21 Oct 2018, 03:10
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