# Discriminant Question

_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_

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

_Mayday_
What happens to the k then?

Staff Emeritus
Gold Member
What happens to the k then?

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

_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?

Staff Emeritus
Gold Member
:rofl:

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

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

a = 1, b = k, c = 8-k?
Sounds good to me

_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.

Staff Emeritus
Gold Member
$$b^2 - {\color{red}2}ac$$
:uhh:

_Mayday_
Dammit, how embaressing.

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

Staff Emeritus
Gold Member
Dammit, how embaressing.

$$b^2 - 4ac$$
$$k^2 - 4(8 - k)$$
$$k^2 - 32 + 4k = 0$$
This line is a good place to stop. Now consider the quadratic equation, what is the condition for there to be real roots?

_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.

Staff Emeritus
Gold Member
For there to be real roots the discriminant need to be equal to or larger than 0.
Correct, so for non-real roots...

_Mayday_
The discriminant must be smaller than 0.

Staff Emeritus
Gold Member
The discriminant must be smaller than 0.
Correct.

_Mayday_

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

??

Staff Emeritus
Gold Member

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

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

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

_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?

Staff Emeritus
Gold Member
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?
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.

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

_Mayday_
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.

Sorry Hoot, I am struggling to understand what that means

Homework Helper
(k-4)(k+8) is equal to 0 when k= -8 or 4. If the product of two numbers is negative, what is must be true of the numbers? (+ times+ = + , - time -= + so ?)

_Mayday_
If the product is negative, then one needs to be negative.

Homework Helper
And the other must be positive. Any number that satisfies k- 4> 0 also satisfies k+ 8> 0. Any number that satisfies k+ 8< 0 also satisfies k-4< 0. What numbers satisy k- 4< 0, k+ 8> 0?

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_Mayday_
-8 < k < 4

Staff Emeritus
Gold Member
-8 < k < 4
Sounds good to me

_Mayday_
Cheers Hoot and Halls!