Discriminant problems

  • #1
discombobulated
41
0
Ok this is the first question:
1) Find the set of values of k for which the equation: f(x)=x2-kx+16 has no real solutions.

This is what I've done:
discriminant < 0

k2-4(16) =0
K2-64=0
k=8

x2-8x+16
(x-4)(x-4)=0
x=4

But, i don't know what to do next now! Please help!

2) The quadratic equation kx2+(4k+1)x +(3k+1)=0 has a repeated root. find the value of k.

I'm not at all sure about this one, but i tried to do something anyway:
(4kx+x)2 -12k2+4k=0
4kx +x -10k = 0
(2k+x)(2x-5) =0

I have no idea what I'm doing, please help!

Thanks!
 

Answers and Replies

  • #2
Atomos
165
0
The first question in part 1 should be +/- 8.

For question 2, the discriminant will be 0. Where you went wrong was including the x. The b term is the cofficient on x, not x and the coefficient and that should be -4k in the end bit of the first step.
 
  • #3
discombobulated
41
0
Thanks, i need to be more careful! I got k= 1/10 for the second question.
And for the first question, i didn't need to work out x? But how do i get the set of values for k? Also, shouldn't it be just -8 because the discriminant is less than 0 for no real solutions?
:confused:
 
  • #4
TD
Homework Helper
1,022
0
discombobulated said:
Thanks, i need to be more careful! I got k= 1/10 for the second question.
And for the first question, i didn't need to work out x? But how do i get the set of values for k? Also, shouldn't it be just -8 because the discriminant is less than 0 for no real solutions?
:confused:
As you say, D has to be < 0 so:
k²-4(16) < 0
K²-64 < 0

Now, watch out:

K² < 64
-8 < k < 8

You see?
 
  • #5
discombobulated
41
0
OH! Thanks!
 
  • #6
ivybond
37
0
Are you sure it's k=1/10?

discombobulated said:
Thanks, i need to be more careful! I got k= 1/10 for the second question.
And for the first question, i didn't need to work out x? But how do i get the set of values for k? Also, shouldn't it be just -8 because the discriminant is less than 0 for no real solutions?
:confused:

I am getting a different answer.
How did you find yours?
Did you check that you get a repeated root x for k=1/10?

I might be wrong too. :blushing:
 
  • #7
discombobulated
41
0
err, what does it mean by 'repeated root', is it like equal real roots?
 
  • #8
Diane_
Homework Helper
393
1
Yes - repeated root and equal roots (not necessarily real) mean the same thing in this case.

If it's any help, I'm getting a -1/2 for k, with the multiple root at x = -1.
 
  • #9
TD
Homework Helper
1,022
0
For k = 1/10, there are no real solutions, if you're referring to question 1 (since that was quoted...)

Diane_ said:
Yes - repeated root and equal roots (not necessarily real) mean the same thing in this case.

If it's any help, I'm getting a -1/2 for k, with the multiple root at x = -1.
Yes, that's correct for question 2.
 
  • #10
discombobulated
41
0
so it's like this:

16x2+1 -12k2-4k=0
4k+1-2k=0
2k+1=0
k=-1/2

but i don't understand the multiple root?!
 
  • #11
TD
Homework Helper
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When the discriminant is 0, we usually say that there is "only 1 solution" (namelijk -b/a), but in fact, there are still 2 solutions, but they 'fall together', it's a 'multiple root' which can be written as (x-a)² = 0 with a the double root.
 
  • #12
paul-martin
27
0
Probebly am i missing something don't know what repeated mean?

x2+4x+2k+3
(x+2)2=-2k-3+4
x=2+-(-2k+1)1/2
 
  • #13
TD
Homework Helper
1,022
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For a quadratic equation of the form [itex]ax^2+bx+c=0[/itex], the two possible solutions are given by

[tex]\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}[/tex]

Now, when the discriminant, [itex]\sqrt {b^2 - 4ac}[/itex], is zero, the two solutions 'fall together", you then have a double root.
 
  • #14
Diane_
Homework Helper
393
1
Geometrically, you know that a quadratic can be graphed as a parabola, with the roots being where the graph crosses the x-axis. Imagine such a parabola opening up, crossing the axis in two places. This would be the case for a quadratic with two real roots.

Now imagine that parabola rising. As it does so, the two roots start to approach each other. When the parabola is just tangent to the axis, the two roots will have merged. This is the case where there is a single root - but it's composed of the two "merged" roots. It's a double root.

Not that it matters, but if you let the parabola continue to rise, it no longer crosses the x-axis. That would be two complex roots.

Does this help any?
 

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