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For which value of k does the following pair...

by Vincen » Tue Sep 12, 2017 11:15 am
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) 4/root(7)

B) root(7)/ 4

C) root(2)

D) -4/root(7)

E) None of these.

The OA is D.

I need help. There are too many variables.
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by EconomistGMATTutor » Fri Oct 20, 2017 3:02 pm
Hello Vincen.

From the second equation we get: (3) $$y^2-k^2=-\left(x+2k\right)^2.$$

Now, from the first equation we can get: (4) $$x^2-2kx+k^2-y^2=8\ <=>x^2-2kx-\left(y^2-k^2\right)=8.$$

Substituiting equation (3) in (4) we will get: (5) $$x^2-2kx+\left(x+2k\right)^2=8\ <=>\ x^2+kx+2k^2-4=0.$$

Now, we want that equation (5) have only one solution. It is equivalent to say that $$\triangle=0.$$

This is to say, $$k^2-4\left(2k^2-4\right)=0\ <=>\ k^2=\frac{16}{7}.$$

The last equation give us two solutions for k, $$k_1=\frac{4}{\sqrt{7}}\ and\ k_2=-\frac{4}{\sqrt{7}}.$$

Now, the solution for the equation (5) is -k/2. We want this solution would be positive, so we have to select the option 2.

The value of k is $$k_2=-\frac{4}{\sqrt{7}}.$$

The correct option is D.

I hope this solution will be enough clear to help you.

Regards.
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