the discrimient

**johnsy123** · March 6th 2010, 10:57 PM

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

find the value for k:
i) two distinct solutions
ii)one solution
iii)no solutions

**Prove It** · March 6th 2010, 11:09 PM

Originally Posted by johnsy123

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

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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