Consider the system:
x + 2z =1
x+2y+4z =1
2x-y+3z=2
For all solutions {x,y,z}, find the range of k such that \(k(x^2-3) > yz \)

- Callisto

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- abb0t

I don't know!! :'(

- nubeer

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

- Callisto

Same here...

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- nubeer

lol sorry guess not much help i am

- Callisto

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

- nubeer

sure , give me a moment.

- anonymous

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

- abb0t

Gaussian elimination method.

- Callisto

|dw:1363361660822:dw|

- Callisto

|dw:1363361796138:dw|

- Callisto

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

- nubeer

k honestly i have done sole me this kind before where we get a who row zero but hard to explain the logic behind it..that whole zero row just made hard to find definite answer.. i normally would do this with the simultaneous method now. so we can get x, y , and z in terms of each other.

- nubeer

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

- Callisto

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

- nubeer

|dw:1363362152973:dw|

- nubeer

so i think we can get value of x , y and z if we suppose any value of z .. according to it , we will get other values.

- Callisto

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

- nubeer

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

- Callisto

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

- nubeer

can try tht

- Callisto

Suppose wolf. is right, then we get \(\frac{z(z+1)}{(-2z^2+2z+1)^2}\)
Put that derivative = 0, z=0 or z=-1

- Callisto

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

- nubeer

hmm thought u were trying to get max and min.

- Callisto

But it won't work?

- nubeer

yeah.. i still feel we using wrong approach.

- anonymous

\[x + 2z =1[1]\]
\[x+2y+4z =1[2]\]
\[2x-y+3z=2[3]\]
In Equation [1]
\[x=1-2z\]
In Equation [2]
\[1-2z+2y+4z=1\]
\[2y+2z=0[4]\]
In Equation [3]
\[2(1-2z)-y+3z=2\]
\[2-4z-y+3z=2\]
\[-y-z=0\]
\[y+z=0[5]\]
Therefore:
\[x=1-2z\]
or
\[x=1+2y\]
\[2z=1-x\]
\[z=\frac{1-x}{2}\]
\[2y=x-1\]
\[y=\frac{x-1}{2}\]
Therefore:
\[yz=\frac{(1-x)(x-1)}{4}\]
In:
\[k(x^2âˆ’3)>yz\]
\[k(x^2-3)>-\frac{(1-x)^2}{4}\]

- anonymous

\[4k(x^2-3)>-(1-x)^2\]
\[4kx^2-12k+(1-2x+x^2)>0\]
\[4kx^2-12k+1-2x+x^2>0\]
\[(4k+1)x^2-2x-(12k-1)>0\]
Using the discriminant:
\[b^2-4ac>0\]
\[4+4(12k-1)(4k+1)>0\]
Solve as quadratic and find the range of values for k.

- Callisto

Almost there, but still not correct.

- mathslover

Man at work.

- AravindG

am I late? :)

- mathslover

I dont' think so.

- anonymous

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

- Callisto

I've tried solving the system above...

- anonymous

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

- Callisto

It shouldn't have a unique solution.

- anonymous

Oh. What level of math is this?

- Callisto

High school

- mathslover

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

- Callisto

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

- anonymous

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

- mathslover

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

- Callisto

No....

- mathslover

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

- Callisto

Sure, take your time :)

- mathslover

I am back to work.

- Callisto

Any guideline?

- mathslover

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

- mathslover

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

- Callisto

No..
diff = different/differentiate?

- mathslover

different.

- anonymous

looks easy
wait

- mathslover

I am getting diff. answers. Wait!

- Callisto

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

- mathslover

That is good.

- shubhamsrg

is it [0, -1/6] ?

- Callisto

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

- shubhamsrg

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

- mathslover

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

- mathslover

is it wrong^ ?

- mathslover

@Callisto ?

- Callisto

It is wrong.

- shubhamsrg

Did you get what I did ?

- shubhamsrg

Coz I am not very confident myself! :P

- Callisto

I didn't get what you did :(

- Callisto

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

- shubhamsrg

I am calling that f(x)
k> f(x)
whatever be the range of f(x), k's range will be opposite of that, i.e. maxima of f(x) is min of k and vice versa.
so, is the ans (0,-1/6) ?

- Callisto

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

- mathslover

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

- Callisto

The inequality itself is > , not >=

- mathslover

k(x^2 - 3)>yz
Put,
x = 1-2z
and y = -z
Then
k{1+4z^2 -4z - 3} > -z^2
=> k(4z^2 - 4z - 2) > -z^2
=> (4k+1)z^2 - 4kz - 2k>0
Now put,
k = 0
you get
z^2 > 0
Which is true ......
So k = 0 satisfies your inequality ....

- mathslover

@Callisto

- anonymous

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

- anonymous

do this,
eliminate 1 from lowest row in matrix
you get two equations of x and z,solve them
then u wil get answers of x,y,z

- mathslover

And MY z = -1/6
have a look:
(4k+1)z^2 - 4kz - 2k>0
Put k =-1/6
We get
z^2 + 2z + 1 >0
OOPS that is true toooooo

- mathslover

So k =-1/6
also satisifies your inequality

- Callisto

What if z=0?

- mathslover

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

- anonymous

yes z=0

- Callisto

@aajugdar I don't mind if you check my matrix work above. I can't find a unique solution to the system.

- mathslover

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

- Callisto

But when z=0, the inequality fails.

- anonymous

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

- anonymous

perform R3- R2/2 on R3 and then again arrange them in equation form,u will get
x+2z =1
and then 3/2x+z =3/2

- mathslover

Well ok in that case we get (by putting in my quadratic inequality in terms of z and k),
k < 1/2
What about that ?

- anonymous

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

- Callisto

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

- anonymous

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

- Callisto

-2 x R1 + R3 -> R3

- Callisto

@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)

- anonymous

nope man
you didn't solve it as it should be solved,you see by makng complete 1 row 0 you wil never get solution
you shouldn't keep a row NULL its a rule

- mathslover

@Callisto very correct

- anonymous

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

- mathslover

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

- Callisto

I am not a man
This system has no unique solution, that's what the determinant of coefficient matrix tell us.

- anonymous

2***

- anonymous

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

- Callisto

@aajugdar
You can try using WolframAlpha, you can't get a unique solution.
@mathslover
Can you guarantee that (4k+1)>0?

- shubhamsrg

k > -(x-1)^2 /(4x^2 -12)
the derivatives are 0 at x=3 and 1
respective values of y are -1/6 (local maxima) and 0(local minima)
so k > -1/6 and k<0

- anonymous

-facewall-

- shubhamsrg

only in that range x>f(x)

- shubhamsrg

I mean k>f(x)

- mathslover

Case - 1 : 4k + 1 > 0 => k>-1/4
Now, that means the inequality has a graph upward. Now for the inequality >0.
b^2 - 4ac < 0
=> k(6k+1) < 0
=> If k< 0 then 6k+1>0 i.e. k>-1/6
And if k>0 then 6k+1 < 0 i.e. k<-1/6 (This option is not possible)
So, k<0 ; k>-1/6 ; and k>-1/4
Take intersection:
k belongs to (-1/6,0)
@Callisto I am thus taking cases

- mathslover

Case -2 :
(4k+1)<0 => k <-1/4 i .e. the graph is bent downwards.
Now for inequality> 0.
b^2 - 4ac > 0
=> k(6k+1) >0
=> k> 0 and k>-1/6 or k<0 and k<-1/6
=> k>0 or k<-1/6
But we took k<-1/4
So, k>0 case is eliminated.
So, k belongs to (-infinity, -1/4)

- mathslover

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

- mathslover

Which obviously consitutes part of real values of z.

- mathslover

So, I said answer should be:
k belongs to (-infinity, -1/4] union (-1/6,0)........
Kindly please tell if I am correct or If I went wrong somewere

- mathslover

Well I agree with you that k = -1/4 and -1/6 do not constitute the solution so modify it
k belongs to (-infinity, -1/4) union (-1/6,0)
But I still don't know that how will I then deal with 4k+1 = 0 case

- Callisto

I'm thinking.. for the inequality, (4k+1)z^2 - 4kz - 2k>0, it doesn't imply that (4k+1) >0 or (4k+1)<0. It can be 4k+1 =0
Oh, you've pointed the problem out.

- mathslover

Think about that

- mathslover

I hope I didn't annoy you?

- Callisto

Nope, it just the exchange of ideas.

- Callisto

*it's

- stamp

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

##### 1 Attachment

- Callisto

The solutions to the system are not numbers, that is it has no unique solution.
The link doesn't wok.

- Callisto

*work

- mathslover

@stamp read the question again ;)

- anonymous

The system is easy to solve for x, y, z. It's just a little tedious. I am in a rush as it's nearly 4 am so I'll give you a brief solution to this problem.
First rearrange x + 2z = 1 to z = (1 - x) / 2
Now you have z in terms of x. Find the value of y in terms of x from the second equation and plug in (1 - x)/2 for z in it. Then plug these y and z values in terms of x in the third equation and solve for x. Then solve for z and then y. You can of course use Gaussian-Elimination Method but that can be annoying at times.
Anyway, after you got the solution for x,y,z, just plug the values in to the inequality and rearrange it to k > yz/(x^2 - 3). And that will give you the range for k. G2g now.
@Callisto

- Callisto

1. There is not unique solution to the system. I have solved it using Gaussian elimination and have already put what I have got above.
2. I have put x, y, z into the inequality (in terms of z), but it just leaves me a range with unknown z.
3. I have tried differentiate z to get min/max, but the values I got (z=0/z=-1) does not match with the answer.

- anonymous

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

- Callisto

Infinitely many solution.

- anonymous

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

- anonymous

I would say range is all real numbers?

- Callisto

No.

- anonymous

no? do you know the answer then?

- Callisto

I know the answer.

- Callisto

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

- anonymous

oh ok got you

- anonymous

Do you know what the smallest solution is for x,y,z such that yz / x^2 - 3 is the smallest possible value? Try to find such a solution.

- Callisto

@genius12 Would you mind checking me steps of solving the system?|dw:1363507811134:dw|

- Callisto

*my

- anonymous

Did you do this through gaussian elimination?

- Callisto

Yes.

- anonymous

Have you tried solving this algebraically? It will be tedious, but could be promising. It's easy to mess up on matrices, but not as much when you solve algebraically.

- anonymous

I know for sure that you will be able to solve this algebraically, and I am pretty confident that you won't get "infinitely many solutions", hopefully.
4:16 AM. Dark room, people sleeping in here, I think I'll head to sleep as well. Good luck Callisto.

- Callisto

x + 2z =1 ---(1)
x+2y+4z =1 --- (2)
2x-y+3z=2 ---(3)
-(1)+(2)
2y + 2z = 0 => y = -z --- (4)
Sub. (4) into (3)
2x - (-z) + 3z =2
x+2z = 1 --- (5) , which is also (1)
Pretty obvious, huh?

- Callisto

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

- Callisto

You'd better sleep now. Good night.

- anonymous

Then there is infinitely many solutions.

- Callisto

What's next?

- anonymous

Wouldn't the system of equations be inconsistent?

- Callisto

The system should be consistent.

- anonymous

Yes, for you to solve the inequality and get the range for k, it should be consistent which it isn't.

- anonymous

@genius12 yes i agree it is inconsistent.

- Callisto

Please point out my mistakes then.

- Callisto

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

- anonymous

I could pick any value for z and get the same three equations. This gives us no way to come up with a range for k. =.=

- Callisto

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

- anonymous

Something wrong with the question? lol

- Callisto

No way! It is a past exam question.

- anonymous

What unit does this question cover?

- Callisto

Matrix and determinant.

- anonymous

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

- Callisto

The final answer - Yes.

- anonymous

Check it? lol

- Callisto

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

- Callisto

When z=-1, (max) => k=min
(x, y, z) = (3, 1, -1)
\[k>\frac{yz}{x^2-3}\]\[k>\frac{(1)(-1)}{3^2-3}\]\[k>-\frac{1}{6}\]
When z = 0, (min) => k = max
(x, y, z) = (1, 0, 0)
\[k<\frac{yz}{x^2-3}\]\[k<\frac{(0)(0)}{1^2-3}\]\[k<0\]
@shubhamsrg Your method works!! Too intuitive for me though :(

- shubhamsrg

I am not too sure how to put my words more properly.
:|
Hey, why don't you try to understand this with the graph, you may use wolfram for the graphing.

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