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

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  • 1 decade ago
    Best 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.

  • Anonymous
    1 decade ago

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

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

  • 3 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.

  • Anonymous
    1 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.

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