I'd really really appreciate if someone could help me figure this out. It's quite easy but I think I'm just overthinking
Find the range of values of K so that Kx^2 - 2Kx - 3K - 12 = 0 has real roots.
Roots being where it cuts the X axis, and real being it actually does cut the X axis. I can try substituting random numbers for K but there must be a quicker way of find the range. Again any help very much appreciated
I'd really really appreciate if someone could help me figure this out. It's quite easy but I think I'm just overthinking
Find the range of values of [b]K[/b] so that [b]K[/b]x^2 - 2[b]K[/b]x - 3[b]K[/b] - 12 = 0 has real roots.
Roots being where it cuts the X axis, and real being it actually does cut the X axis. I can try substituting random numbers for [b]K[/b] but there must be a quicker way of find the range. Again any help very much appreciated
You can use the discriminant http://en.wikipedia.org/wiki/Discriminant
Which will be positive for all cases where the quadratic has two real roots. If you substitute in the coefficients of each term and let the discriminant > 0, you can solve for K.
You can use the discriminant http://en.wikipedia.org/wiki/Discriminant
Which will be positive for all cases where the quadratic has two real roots. If you substitute in the coefficients of each term and let the discriminant > 0, you can solve for K.
tornados2111Find the range of values of K so that Kx^2 - 2Kx - 3K - 12 = 0 has real roots.
Not sure if that means that the equation must have 2 real roots for every K, if not then it's discriminant >= 0.
[quote=tornados2111]Find the range of values of [b]K[/b] so that [b]K[/b]x^2 - 2[b]K[/b]x - 3[b]K[/b] - 12 = 0 has real roots.[/quote]
Not sure if that means that the equation must have 2 real roots for every [b]K[/b], if not then it's discriminant >= 0.
Boar If you substitute in the coefficients of each term and let the discriminant > 0, you can solve for K.
EDIT: Nvm, made a mistake
[quote=Boar] If you substitute in the coefficients of each term and let the discriminant > 0, you can solve for K.[/quote]
EDIT: Nvm, made a mistake
What I'm confused about is to complete the formula "b^2 - 4ac", I need to find a,b and c.
Are they K, -2K and -3K - 12 or are they 1, -2 and -3 - 12. In other words, do I need to include the K?
To me, it only makes sense that the K be included so I used the formula and got "16k^2 + 48K >= 0"
What am I supposed to do with that?
What I'm confused about is to complete the formula "b^2 - 4ac", I need to find a,b and c.
Are they K, -2K and -3K - 12 or are they 1, -2 and -3 - 12. In other words, do I need to include the K?
To me, it only makes sense that the K be included so I used the formula and got "16k^2 + 48K >= 0"
What am I supposed to do with that?
16k^2 + 48k >= 0
k^2 + 3k >= 0
k(k+3) >= 0
You can split that up into
k >= 0
and
k + 3 >= 0
k >= -3
k >= 0 includes the other one, so that's the range
this is wrong
16k^2 + 48k >= 0
k^2 + 3k >= 0
k(k+3) >= 0
[s]You can split that up into
k >= 0
and
k + 3 >= 0
k >= -3
k >= 0 includes the other one, so that's the range[/s]
this is wrong
(a)x^2 + (b) x + c
(K)x^2 + (-2K)x + (-3K - 12)
...so yes, include the K.
Then solve for K.
(a)x^2 + (b) x + c
(K)x^2 + (-2K)x + (-3K - 12)
...so yes, include the K.
Then solve for K.
Ahhhhh finally I get it. Thanks everyone for their help!
Ahhhhh finally I get it. Thanks everyone for their help!
Whoops I lied, you actually have to set up an interval chart like this:
............ k < -3 .. -3 < k < 0 ... k > 0
k ............ (-) ............ (-) .......... (+)
k + 3 ...... (-) ........... (+) .......... (+)
f(k) ........ (+) ............ (-) .......... (+)
Which gives f(k) >= 0 when k <= -3 or k >= 0. You could also draw it out and look at it, or just do it in your head since it's just a parabola.
Sorry for confusing you, it's been a while.
not sure if the spacing works but hopefully you get the idea
Whoops I lied, you actually have to set up an interval chart like this:
............ k < -3 .. -3 < k < 0 ... k > 0
k ............ (-) ............ (-) .......... (+)
k + 3 ...... (-) ........... (+) .......... (+)
f(k) ........ (+) ............ (-) .......... (+)
Which gives f(k) >= 0 when k <= -3 or k >= 0. You could also draw it out and look at it, or just do it in your head since it's just a parabola.
Sorry for confusing you, it's been a while.
not sure if the spacing works but hopefully you get the idea
Yeah, that makes sense but we weren't taught this so I think the other way is fine for us. I don't see why the other way won't work, at least they work for all of problems we are solving. But thanks for making that clear in case!
Yeah, that makes sense but we weren't taught this so I think the other way is fine for us. I don't see why the other way won't work, at least they work for all of problems we are solving. But thanks for making that clear in case!