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I don't know!! :'(

hmm i can find values of x , y and z but don;t know what to do after wards with them

Same here...

lol sorry guess not much help i am

Wait, can you find the numerical values for x, y, z?

sure , give me a moment.

what about first finding the solution set of the system using algebra matrix

Gaussian elimination method.

|dw:1363361660822:dw|

|dw:1363361796138:dw|

That's what I can do and that's why I ask if you can find the numerical value for {x,y,z}

if i solve simultaneously, i got
z = -y and
x = 1 -2z

Yes, I got the same too, but how can we utilize it to find range of k?

|dw:1363362152973:dw|

\[k[(1-2z)^2-3]>-z^2\]\[k>\frac{-z^2}{(1-2z)^2-3}\]Hmm..

hmm got same thing.. after this i did the long division.. but lol not possible way out

I'm thinking of differetiating the fraction to find the min/max/range..

can try tht

Oh wait.. Why would I differentiate that!!!

hmm thought u were trying to get max and min.

But it won't work?

yeah.. i still feel we using wrong approach.

Almost there, but still not correct.

Man at work.

am I late? :)

I dont' think so.

Can you just solve for x,y,and z (using a system of equations) then plug the variables back in?

I've tried solving the system above...

I ended up with x=1, y=0, z=0.
So, \[k(1 ^{2}-3)>(0)(0)\]
=>\[k(-2)>0\]
=>k<0

It shouldn't have a unique solution.

Oh. What level of math is this?

High school

k belongs to ( - infinity , -1/4 ] union (-1/6 , 0 )
?

No, almost there. Can you tell us how you get there?

wait, how is k<0 unique? (-inf, 0)

k belongs to ( - infinity , -1/4 ] union (-1/6 , 0 ] ?

No....

can I post here after 30 minutes please? I have to go for dinner ?

Sure, take your time :)

I am back to work.

Any guideline?

Well first of all I solved for x and then put it into the eqn and then took diff. conditions .

k belongs to ( - infinity , -1/4 ] union [-1/6 , 0 ] ?

No..
diff = different/differentiate?

different.

looks easy
wait

I am getting diff. answers. Wait!

I have some patience, so I keep waiting, and trying :)

That is good.

is it [0, -1/6] ?

Very very close!!!
How do you get it??

-(x-1)^2 /(4x^2 -12)
tried to work with this :|

k belongs to [-1/6,0] union (-infinity ,1/4] .. i am still getting it

is it wrong^ ?

It is wrong.

Did you get what I did ?

Coz I am not very confident myself! :P

I didn't get what you did :(

Did you start from this:\[k(x^2-3)>-\frac{(1-x)^2}{4}\]?

The answer is correct, but I don't understand your explanation.

But aren't 0 and -1/6 satisfying the equation ?

The inequality itself is > , not >=

lol
x=1
btw I was having dinner
z=0
and y =-1/2

So k =-1/6
also satisifies your inequality

What if z=0?

Well if z = 0
then go for it:
k < 1/2

yes z=0

Well the solutions given by you and shubhamsrg have many cases left out .....

But when z=0, the inequality fails.

put them in equation dude,the values are correct
ok hold on,i wil tell ya

solve them,u will get values of x and z,from that values,find value of y by putting it in equation 2

@mathslover That is not even the answer
@aajugdar Check this
|dw:1363447547599:dw|

what operation did you perform on 3rd row????

-2 x R1 + R3 -> R3

@mathlover When I learnt inequality, my teacher told me that if
(1)
ax^2+bx+c>0, where a>0,
then the solution would be like (alpha)(beta)
(2)
ax^2+bx+c<0, where a>0,
then the solution would be like (alpha)

@Callisto very correct

remember
3 2 unknowns,2 equations,3 unknowns 3 eqautions.

(4k+1)z^2 - 4kz - 2k>0
I used that (please have a look at my solution)

2***

well,solve it correctly,it has a solution
I found it

-facewall-

only in that range x>f(x)

I mean k>f(x)

Case - 3:
4k+1 = 0
=> k = -1/4
So,
inequality reduces to
z>-1/2

Which obviously consitutes part of real values of z.

Think about that

I hope I didn't annoy you?

Nope, it just the exchange of ideas.

*it's

it seems the solutions are not numbers, but rather variables (see attachment)
also, http://tutorial.math.lamar.edu/Classes/Alg/AugmentedMatrixII.aspx

*work

@stamp read the question again ;)

So does the system have infinite solutions or a finite number of solutions?

Infinitely many solution.

If it has infinitely many solutions then which solution gives the smallest value for yz / x^2-3?

I would say range is all real numbers?

No.

no? do you know the answer then?

I know the answer.

I don't know how to get the answer though.

oh ok got you

*my

Did you do this through gaussian elimination?

Yes.

If it is not infinitely many solutions, then it will be no solution.

You'd better sleep now. Good night.

Then there is infinitely many solutions.

What's next?

Wouldn't the system of equations be inconsistent?

The system should be consistent.

Please point out my mistakes then.

FYI: http://www.wolframalpha.com/input/?i=x+%2B+2z+%3D1+and+x%2B2y%2B4z+%3D1+and+2x-y%2B3z%3D2

That's why we need to think how to get a way to solve it.

Something wrong with the question? lol

No way! It is a past exam question.

What unit does this question cover?

Matrix and determinant.

If it's a past exam, isn't there a given solution?

The final answer - Yes.

Check it? lol

I know what the final answer is. I just don't know how to get the final answer.