# Discriminant Question

1. Apr 23, 2008

### _Mayday_

[SOLVED] Discriminant Question

The Equation:

$x^2 + kx + 8 = k$

has no real roots

(a) Show that $k$ satisfies $k^2 + 4k - 32 (is.smaller.than) 0$

(b) Hence find the set of possible values of $k$

Attempt

(a) First I will put it in the form:
$ax^2 + bx + c = 0$
$x^2 + kx + 8 = k$

a = 1, b = k, c = 8

Now I will put this into the discriminant.

$b^2 - 2ac (is.smaller.than) k$
$k^2 - 16 (is.smaller.than) k$

Where do I go from here??

Thanks for any help!

_Mayday_

2. Apr 23, 2008

### Hootenanny

Staff Emeritus
You have said that you were going to put it into canonical form, but haven't done it! Your RHS should be zero.

3. Apr 23, 2008

### _Mayday_

What happens to the k then?

4. Apr 23, 2008

### Hootenanny

Staff Emeritus
$$x^2+kx+8-k=0$$

5. Apr 23, 2008

### _Mayday_

:rofl:

$$x^2 + kx + 8 - k$$

Ok, but now how do I know which bit is which?

a = 1, b = k, c = 8-k?

6. Apr 23, 2008

### Hootenanny

Staff Emeritus
Sounds good to me

7. Apr 23, 2008

### _Mayday_

$$b^2 - 2ac$$
$$k^2 - 2(8 - k)$$
$$k^2 - 16 + 2k = 0$$
$$k^2 - 2k = - 16$$

Where do I go from here? I need to prove that the LHS is smaller than 0.

8. Apr 23, 2008

### Hootenanny

Staff Emeritus
:uhh:

9. Apr 23, 2008

### _Mayday_

Dammit, how embaressing.

$$b^2 - 4ac$$
$$k^2 - 4(8 - k)$$
$$k^2 - 32 + 4k = 0$$
$$k^2 + 4k = 32$$

10. Apr 23, 2008

### Hootenanny

Staff Emeritus
This line is a good place to stop. Now consider the quadratic equation, what is the condition for there to be real roots?

11. Apr 23, 2008

### _Mayday_

The question says there are no real roots. For there to be real roots the discriminant need to be equal to or larger than 0.

12. Apr 23, 2008

### Hootenanny

Staff Emeritus
Correct, so for non-real roots...

13. Apr 23, 2008

### _Mayday_

The discriminant must be smaller than 0.

14. Apr 23, 2008

### Hootenanny

Staff Emeritus
Correct.

15. Apr 23, 2008

### _Mayday_

$$k^2 - 32 + 4k (smaller.than) 0$$

??

16. Apr 23, 2008

### Hootenanny

Staff Emeritus
Indeed it is, which is what you were asked to prove.

P.S. Latex allows use of '<' symbols.

17. Apr 23, 2008

### _Mayday_

Haha what a relief! Never knew > worked!

Now to find possible values I simply factorise?

$$(k - 4)(k + 8)$$

So the possible values are 4 and -8?

18. Apr 23, 2008

### Hootenanny

Staff Emeritus
No these values provide you with the range of an interval. Now you need to decide whether the segment you want (i.e. the set of numbers which yield complex roots) lies within this interval or outside it.

19. Apr 23, 2008

### _Mayday_

Hoot, I gotta dash. I'll be back on in an hour or so, see you if you're on. Thanks for your help upto now though!

20. Apr 23, 2008

### _Mayday_

Sorry Hoot, I am struggling to understand what that means