# Find the range of values of k for which curve y=16/x intersects the straight line y=k-x at two distinct points?

Find the range of values of k for which curve y=16/x intersects the straight line y=k-x at two distinct points.

The correct answer is k < -8, k > 8. My current answer is k < -8, k < 8 which is close. I have no idea which formula to use since it isn't mentioned in the question. I used b^2-4ac < 0 to match with the correct answer.

Please explain.

Thanks.

### 4 Answers

- Walid JLv 71 decade agoBest Answer
16/x = k - x

16 = kx - x^2

so x^2 -kx +16 = 0

b^2-4ac = k^2 - 64 > 0 (in order to have two distinct roots)

k^2 > 64

take sqrt for both sides and not that sqrt(k^2) = |k|

u get:

| k | > 8

so either k > 8 or k < -8

was this helpful?!! i hope that.

- Anonymous1 decade ago
for intersect 16/x = k-x --->x^2 - kx + 16 = 0

discriminant = k^2 - 64 >0 -----> k< - 8 or k >8

- pinkleyLv 43 years ago
you need to locate the place they intersect, via substituting the 2d equation into the 1st. this provides you with some algebra decrease than a sq. root sign which could in ordinary terms be solved via inequality. because of the fact the equation has no genuine roots, then the equation must be < 0.

- Anonymous1 decade ago
Where did u get the 'real answer' from. (back of a textbook). I believe the

'real answer' is incorrect since the (k < -8) is within the domain of (k < 8)

In any case, that solution appears false.