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

find the value for k:

i) two distinct solutions

ii)one solution

iii)no solutions

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- Mar 6th 2010, 10:57 PM #1

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- Jun 2009
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- Mar 6th 2010, 11:09 PM #2
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 $\displaystyle k > 1$.