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    the discrimient

    $\displaystyle x^2+4kx+4=0$

    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
    $\displaystyle x^2+4kx+4=0$

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

    For no solutions: $\displaystyle \Delta < 0$.

    For one solution: $\displaystyle \Delta = 0$.

    For two solutions: $\displaystyle \Delta > 0$.


    Let's start with where $\displaystyle \Delta = 0$

    $\displaystyle b^2 - 4ac = 0$

    $\displaystyle (4k)^2 - 4(1)(4) = 0$

    $\displaystyle 16k^2 - 16 = 0$

    $\displaystyle 16k^2 = 16$

    $\displaystyle k^2 = 1$

    $\displaystyle k = \pm 1$.


    So for one solution, $\displaystyle k = -1$ or $\displaystyle k = 1$.


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

    Have you heard of the absolute value function?

    $\displaystyle |x| = x$ if $\displaystyle x > 0$ and $\displaystyle |x| = -x$ if $\displaystyle x < 0$.

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


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

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

    This is particularly useful if dealing with squares in inequalities.


    Now back to the problem.

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


    $\displaystyle \Delta < 0$

    $\displaystyle b^2 - 4ac < 0$

    $\displaystyle (4k)^2 - 4(1)(4) < 0$

    $\displaystyle 16k^2 - 16 < 0$

    $\displaystyle 16k^2 < 16$

    $\displaystyle k^2 < 1$.

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

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

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

    $\displaystyle |k| < 1$.


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

    This means $\displaystyle -1 < k < 1$. 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, $\displaystyle -1 < k < 1$.



    Finally, for 2 solutions:

    $\displaystyle \Delta > 0$

    $\displaystyle b^2 - 4ac > 0$

    $\displaystyle (4k)^2 - 4(1)(4) > 0$

    $\displaystyle 16k^2 - 16 > 0$

    $\displaystyle 16k^2 > 16$

    $\displaystyle k^2 > 1$

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

    $\displaystyle |k| > 1$.


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

    This means $\displaystyle k < -1$ or $\displaystyle k > 1$.

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


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



    To summarise:

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

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

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