• Support PF! Buy your school textbooks, materials and every day products Here!

Discriminant Question

  • Thread starter _Mayday_
  • Start date
795
0
[SOLVED] Discriminant Question

The Equation:

[itex]x^2 + kx + 8 = k[/itex]

has no real roots

(a) Show that [itex]k[/itex] satisfies [itex] k^2 + 4k - 32 (is.smaller.than) 0[/itex]

(b) Hence find the set of possible values of [itex]k[/itex]


Attempt

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

a = 1, b = k, c = 8

Now I will put this into the discriminant.

[itex]b^2 - 2ac (is.smaller.than) k[/itex]
[itex]k^2 - 16 (is.smaller.than) k[/itex]

Where do I go from here??

Thanks for any help!


_Mayday_
 

Answers and Replies

Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
(a) First I will put it in the form:
[itex]ax^2 + bx + c = 0[/itex]
[itex]x^2 + kx + 8 = k[/itex]

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.
 
795
0
What happens to the k then?
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
795
0
:blushing::rofl:

[tex]x^2 + kx + 8 - k[/tex]

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

a = 1, b = k, c = 8-k?
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
:blushing::rofl:

[tex]x^2 + kx + 8 - k[/tex]

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

a = 1, b = k, c = 8-k?
Sounds good to me :biggrin:
 
795
0
[tex]b^2 - 2ac[/tex]
[tex]k^2 - 2(8 - k)[/tex]
[tex]k^2 - 16 + 2k = 0[/tex]
[tex]k^2 - 2k = - 16[/tex]

Where do I go from here? I need to prove that the LHS is smaller than 0.
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
795
0
Dammit, how embaressing.

[tex]b^2 - 4ac[/tex]
[tex]k^2 - 4(8 - k)[/tex]
[tex]k^2 - 32 + 4k = 0[/tex]
[tex]k^2 + 4k = 32[/tex]
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
Dammit, how embaressing.

[tex]b^2 - 4ac[/tex]
[tex]k^2 - 4(8 - k)[/tex]
[tex]k^2 - 32 + 4k = 0[/tex]
This line is a good place to stop. Now consider the quadratic equation, what is the condition for there to be real roots?
 
795
0
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.
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
795
0
The discriminant must be smaller than 0.
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
795
0
So then my answer is:

[tex]k^2 - 32 + 4k (smaller.than) 0[/tex]

??
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
So then my answer is:

[tex]k^2 - 32 + 4k (smaller.than) 0[/tex]

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

P.S. Latex allows use of '<' symbols.
 
795
0
Haha what a relief! Never knew > worked!

Now to find possible values I simply factorise?

[tex](k - 4)(k + 8)[/tex]

So the possible values are 4 and -8?
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
Haha what a relief! Never knew > worked!

Now to find possible values I simply factorise?

[tex](k - 4)(k + 8)[/tex]

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.
 
795
0
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!
 
795
0
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 :confused:
 
HallsofIvy
Science Advisor
Homework Helper
41,734
893
(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 ???)
 
795
0
If the product is negative, then one needs to be negative.
 
HallsofIvy
Science Advisor
Homework Helper
41,734
893
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?
 
Last edited by a moderator:
795
0
-8 < k < 4
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6

Related Threads for: Discriminant Question

  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
6
Views
1K
Replies
7
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
2
Views
653
Replies
7
Views
736
Replies
35
Views
10K
Replies
1
Views
1K
Top