Homework Help: Discriminant problems

discombobulated · Sep 26, 2005

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!

Atomos · Sep 26, 2005

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.

discombobulated · Sep 26, 2005

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?

TD · Sep 26, 2005

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?

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?

discombobulated · Sep 26, 2005

OH! Thanks!

ivybond · Sep 27, 2005

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?

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.

discombobulated · Sep 27, 2005

err, what does it mean by 'repeated root', is it like equal real roots?

Diane_ · Sep 27, 2005

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.

TD · Sep 27, 2005

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.

discombobulated · Sep 27, 2005

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?!

TD · Sep 27, 2005

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.

paul-martin · Sep 27, 2005

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

TD · Sep 27, 2005

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.

Diane_ · Sep 27, 2005

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?