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Math Help - the discrimient

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    the discrimient

    x^2+4kx+4=0

    find the value for k:
    i) two distinct solutions
    ii)one solution
    iii)no solutions
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    Quote Originally Posted by johnsy123 View Post
    x^2+4kx+4=0

    find the value for k:
    i) two distinct solutions
    ii)one solution
    iii)no solutions
    Ok, so you should know that the discriminant \Delta = b^2 - 4ac determines how many solutions there are going to be.

    For no solutions: \Delta < 0.

    For one solution: \Delta = 0.

    For two solutions: \Delta > 0.


    Let's start with where \Delta = 0

    b^2 - 4ac = 0

    (4k)^2 - 4(1)(4) = 0

    16k^2 - 16 = 0

    16k^2 = 16

    k^2 = 1

    k = \pm 1.


    So for one solution, k = -1 or k = 1.


    It's a little harder working out the values of k for which there are no solutions or two solutions.

    Have you heard of the absolute value function?

    |x| = x if x > 0 and |x| = -x if x < 0.

    In other words, |x| represents the SIZE of x (if you were to draw a number line and measure the distance from 0 to x).


    Now, since \sqrt{x^2} = \pm x, that means that \sqrt{x^2} is something that has SIZE x.

    This means \sqrt{x^2} = |x|.

    This is particularly useful if dealing with squares in inequalities.


    Now back to the problem.

    Let's find the value of k for which there are not any solutions.


    \Delta < 0

    b^2 - 4ac < 0

    (4k)^2 - 4(1)(4) < 0

    16k^2 - 16 < 0

    16k^2 < 16

    k^2 < 1.

    Now to undo this square, we need to take the square root:

    \sqrt{k^2} < \sqrt{1}

    But \sqrt{k^2} = |k|, so

    |k| < 1.


    So, this means that the "size" of k has to be less than 1.

    This means -1 < k < 1. You can verify this by drawing a number line (the distance of any of the points in this region from 0 is less than 1).

    So for there to be no solutions, -1 < k < 1.



    Finally, for 2 solutions:

    \Delta > 0

    b^2 - 4ac > 0

    (4k)^2 - 4(1)(4) > 0

    16k^2 - 16 > 0

    16k^2 > 16

    k^2 > 1

    \sqrt{k^2} > \sqrt{1}

    |k| > 1.


    So the size of k has to be greater than 1.

    This means k < -1 or k > 1.

    Once again, you can check this with a number line. The distance from any point < -1 or > 1 to 0 is obviously going to be > 1.


    Therefore, for 2 solutions: k < -1 or k > 1.



    To summarise:

    0 solutions: |k| < 1, which is the same as -1 < k < 1.

    1 solution: |k| = 1, which is the same as k = -1 or k = 1.

    2 solutions: |k| > 1, which is the same as k < -1 or k > 1.
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