discriminant problems

discombobulated

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

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

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

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

OH! Thanks!

ivybond

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

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

Diane_

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

For k = 1/10, there are no real solutions, if you're referring to question 1 (since that was quoted...)

Yes, that's correct for question 2.

discombobulated

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

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

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

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_

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?