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Gughanath
Oct2-04, 08:02 AM
what are the range of values of k that gives the equation (k+1)x^2+4kx+9=0 ...I work it out :confused: ...please help

arildno
Oct2-04, 08:05 AM
Welcome to PF!
I think you have omitted something from your text; what was that?

Gughanath
Oct2-04, 08:07 AM
ups sorry...it should say at the end...that gives the equation no real roots

arildno
Oct2-04, 08:14 AM
And what do you think that means?

shmoe
Oct2-04, 08:16 AM
Hi, you can use the quadratic formula to find the roots of your equation. Forcing no real roots is equivalent to forcing the discriminant (the part under the root sign) to be negative. This will give you conditions on k that you're looking for.

ps. you have to make a restriction on k to guarantee that your equation is a quadratic and not linear. what is this restriction?

Gughanath
Oct2-04, 08:41 AM
i cant use the quadrativ formula for this euqation..when i work out the bracket i get
x^2+x^2k+4kx+9=0 ...i have no idea how to use the discrimant in this case..please help

arildno
Oct2-04, 08:45 AM
Let a=k+1, b=4k, c=9.
Then your equation looks like
ax^2+bx+c=0
Can you solve that one?

Gughanath
Oct2-04, 08:47 AM
i know that the discrimiant has to be smaller than 0....

arildno
Oct2-04, 08:50 AM
And what is the discriminant, expressed with a,b and c?

Gughanath
Oct2-04, 08:52 AM
no i am still confused

arildno
Oct2-04, 08:54 AM
What is your problem?

Gughanath
Oct2-04, 08:55 AM
could you please just show me how i work out this questions?

Gughanath
Oct2-04, 08:56 AM
because then i will undertand

arildno
Oct2-04, 08:59 AM
I ask you again:
Given the equation:
ax^2+bx+c=0
What is the discriminant?

Gughanath
Oct2-04, 09:02 AM
16k^2-(4[k+1]*9)<0 that becomes 16k^2-36k-36<0

shmoe
Oct2-04, 09:03 AM
16k^2-(4[k+1]*9)<0 that becomes 16k^2-36k-36<0

Good, now can you find the values of k that satisfy this new inequality?

arildno
Oct2-04, 09:15 AM
Very good!
Here's a hint:
In order to find the range of k-values your after,
1. find the zeros in your discriminant.
That is, solve the equation for k:
16k^{2}-36k-36=0
2. You weren't interested in the k-values for which the discriminant was zero, but the k-values for which the discriminant is less than zero.
But you should figure out for yourself that those values must lie between the two values found in 1.

Gughanath
Oct2-04, 09:15 AM
i see that i have to factoris ethe equation in the disriminant now..but i cant find the right numbers

arildno
Oct2-04, 09:19 AM
i see that i have to factoris ethe equation in the disriminant now..but i cant find the right numbers
Quite true!
Look at my previous post for hints.

Gughanath
Oct2-04, 09:20 AM
please...i have no idea how to continue...

arildno
Oct2-04, 09:23 AM
Well, what values of k solves:
16k^{2}-36k-36=0 ?

(Note: You were asked to find the values of k so that the discriminant is less than zero, not zero, but finding the zeroes is a good start)

Gughanath
Oct2-04, 09:26 AM
k..thanks for your help...

arildno
Oct2-04, 09:38 AM
Now, having found the k-values yielding zero discriminant, you should be able to write the discrimanant as:
16k^{2}-36k-36=16(k-3)(k+\frac{3}{4})
What must then the k-interval be which yields negative discriminant?

Gughanath
Oct2-04, 04:35 PM
how did u factorise that?

arildno
Oct2-04, 04:51 PM
It is the roots of equation gained by setting the discriminant equal to 0 (that is, -3/4 and 3 are the roots)

Gughanath
Oct3-04, 03:40 AM
ooo (: im so.....now i undertand it...the discriminant is 16k^2+36k-36..simplified it becomes 16[k^2 - (9/4)k - 9/4]...-3 and +3/4 multiplied give -9/4 and when added they also give -9/4...so these are these are the roots..factorised the equation becomes 16[(k-3)(k+3/4)]=0..the values of k here is 3 or -3/4...but the equation must be lower than 0 for 16[(k-3)(k+3/4)]<0 the range of the values of k are in -3/4<k<3...is that right?

Gughanath
Oct3-04, 03:41 AM
thanks alot for your help everyone...sorry for bothering you..i was so busy yesterday my brain wasant running properly...lol

arildno
Oct3-04, 06:09 AM
You're right about the interval;I wouldn't agree to what you said about your brain, though..

TenaliRaman
Oct3-04, 02:20 PM
very good so any k that satisfies 16k^2-36k-36<0 is the k we are looking for ..
lets divide throughout by 4 to make the eqn simple
we get 4k^2-9k-9<0
i can write this as,
(4k+1)(k-3)<0

what does this imply?
(Note : if x*y < 0 it means multiplication of x and y must be negative)

-- AI

TenaliRaman
Oct3-04, 02:22 PM
damn ! i did not realise this thread had 2 pages!!
umm please delete my current and last post as they are totally irrelevant now!

-- AI

arildno
Oct3-04, 02:52 PM
In addition, it should be
(4k+3)(k-3)<0
Drown in shame..:wink:

TenaliRaman
Oct3-04, 03:43 PM
*drowns himself in a bowl of red wine*

-- AI

arildno
Oct3-04, 03:47 PM
TENALIRAMAN!!
Are you still with us??
Get out of that bowl at once, I'm sorry I led you on to that!! :cry:

TenaliRaman
Oct4-04, 02:29 AM
ssh arildno,
i am waiting for some baywatch girl to give me CPR

-- AI

Spectre5
Oct5-04, 07:52 PM
lol, it was all setup