Callisto
  • Callisto
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 \(k(x^2-3) > yz \)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
abb0t
  • abb0t
I don't know!! :'(
nubeer
  • nubeer
hmm i can find values of x , y and z but don;t know what to do after wards with them
Callisto
  • Callisto
Same here...

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nubeer
  • nubeer
lol sorry guess not much help i am
Callisto
  • Callisto
Wait, can you find the numerical values for x, y, z?
nubeer
  • nubeer
sure , give me a moment.
anonymous
  • anonymous
what about first finding the solution set of the system using algebra matrix
abb0t
  • abb0t
Gaussian elimination method.
Callisto
  • Callisto
|dw:1363361660822:dw|
Callisto
  • Callisto
|dw:1363361796138:dw|
Callisto
  • 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
  • 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
  • nubeer
if i solve simultaneously, i got z = -y and x = 1 -2z
Callisto
  • Callisto
Yes, I got the same too, but how can we utilize it to find range of k?
nubeer
  • nubeer
|dw:1363362152973:dw|
nubeer
  • 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
  • Callisto
\[k[(1-2z)^2-3]>-z^2\]\[k>\frac{-z^2}{(1-2z)^2-3}\]Hmm..
nubeer
  • nubeer
hmm got same thing.. after this i did the long division.. but lol not possible way out
Callisto
  • Callisto
I'm thinking of differetiating the fraction to find the min/max/range..
nubeer
  • nubeer
can try tht
Callisto
  • 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
  • Callisto
Oh wait.. Why would I differentiate that!!!
nubeer
  • nubeer
hmm thought u were trying to get max and min.
Callisto
  • Callisto
But it won't work?
nubeer
  • nubeer
yeah.. i still feel we using wrong approach.
anonymous
  • 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: \[k(x^2−3)>yz\] \[k(x^2-3)>-\frac{(1-x)^2}{4}\]
anonymous
  • 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
  • Callisto
Almost there, but still not correct.
mathslover
  • mathslover
Man at work.
AravindG
  • AravindG
am I late? :)
mathslover
  • mathslover
I dont' think so.
anonymous
  • anonymous
Can you just solve for x,y,and z (using a system of equations) then plug the variables back in?
Callisto
  • Callisto
I've tried solving the system above...
anonymous
  • anonymous
I ended up with x=1, y=0, z=0. So, \[k(1 ^{2}-3)>(0)(0)\] =>\[k(-2)>0\] =>k<0
Callisto
  • Callisto
It shouldn't have a unique solution.
anonymous
  • anonymous
Oh. What level of math is this?
Callisto
  • Callisto
High school
mathslover
  • mathslover
k belongs to ( - infinity , -1/4 ] union (-1/6 , 0 ) ?
Callisto
  • Callisto
No, almost there. Can you tell us how you get there?
anonymous
  • anonymous
wait, how is k<0 unique? (-inf, 0)
mathslover
  • mathslover
k belongs to ( - infinity , -1/4 ] union (-1/6 , 0 ] ?
Callisto
  • Callisto
No....
mathslover
  • mathslover
can I post here after 30 minutes please? I have to go for dinner ?
Callisto
  • Callisto
Sure, take your time :)
mathslover
  • mathslover
I am back to work.
Callisto
  • Callisto
Any guideline?
mathslover
  • mathslover
Well first of all I solved for x and then put it into the eqn and then took diff. conditions .
mathslover
  • mathslover
k belongs to ( - infinity , -1/4 ] union [-1/6 , 0 ] ?
Callisto
  • Callisto
No.. diff = different/differentiate?
mathslover
  • mathslover
different.
anonymous
  • anonymous
looks easy wait
mathslover
  • mathslover
I am getting diff. answers. Wait!
Callisto
  • Callisto
I have some patience, so I keep waiting, and trying :)
mathslover
  • mathslover
That is good.
shubhamsrg
  • shubhamsrg
is it [0, -1/6] ?
Callisto
  • Callisto
Very very close!!! How do you get it??
shubhamsrg
  • shubhamsrg
-(x-1)^2 /(4x^2 -12) tried to work with this :|
mathslover
  • mathslover
k belongs to [-1/6,0] union (-infinity ,1/4] .. i am still getting it
mathslover
  • mathslover
is it wrong^ ?
mathslover
  • mathslover
@Callisto ?
Callisto
  • Callisto
It is wrong.
shubhamsrg
  • shubhamsrg
Did you get what I did ?
shubhamsrg
  • shubhamsrg
Coz I am not very confident myself! :P
Callisto
  • Callisto
I didn't get what you did :(
Callisto
  • Callisto
Did you start from this:\[k(x^2-3)>-\frac{(1-x)^2}{4}\]?
shubhamsrg
  • 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
  • Callisto
The answer is correct, but I don't understand your explanation.
mathslover
  • mathslover
But aren't 0 and -1/6 satisfying the equation ?
Callisto
  • Callisto
The inequality itself is > , not >=
mathslover
  • 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
  • mathslover
@Callisto
anonymous
  • anonymous
lol x=1 btw I was having dinner z=0 and y =-1/2
anonymous
  • 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
  • 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
  • mathslover
So k =-1/6 also satisifies your inequality
Callisto
  • Callisto
What if z=0?
mathslover
  • mathslover
Well if z = 0 then go for it: k < 1/2
anonymous
  • anonymous
yes z=0
Callisto
  • Callisto
@aajugdar I don't mind if you check my matrix work above. I can't find a unique solution to the system.
mathslover
  • mathslover
Well the solutions given by you and shubhamsrg have many cases left out .....
Callisto
  • Callisto
But when z=0, the inequality fails.
anonymous
  • anonymous
put them in equation dude,the values are correct ok hold on,i wil tell ya
anonymous
  • 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
  • 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
  • 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
  • Callisto
@mathslover That is not even the answer @aajugdar Check this |dw:1363447547599:dw|
anonymous
  • anonymous
what operation did you perform on 3rd row????
Callisto
  • Callisto
-2 x R1 + R3 -> R3
Callisto
  • 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
  • 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
  • mathslover
@Callisto very correct
anonymous
  • anonymous
remember 3 2 unknowns,2 equations,3 unknowns 3 eqautions.
mathslover
  • mathslover
(4k+1)z^2 - 4kz - 2k>0 I used that (please have a look at my solution)
Callisto
  • Callisto
I am not a man This system has no unique solution, that's what the determinant of coefficient matrix tell us.
anonymous
  • anonymous
2***
anonymous
  • anonymous
well,solve it correctly,it has a solution I found it
Callisto
  • Callisto
@aajugdar You can try using WolframAlpha, you can't get a unique solution. @mathslover Can you guarantee that (4k+1)>0?
shubhamsrg
  • 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
  • anonymous
-facewall-
shubhamsrg
  • shubhamsrg
only in that range x>f(x)
shubhamsrg
  • shubhamsrg
I mean k>f(x)
mathslover
  • 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
  • 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
  • mathslover
Case - 3: 4k+1 = 0 => k = -1/4 So, inequality reduces to z>-1/2
mathslover
  • mathslover
Which obviously consitutes part of real values of z.
mathslover
  • 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
  • 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
  • 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
  • mathslover
Think about that
mathslover
  • mathslover
I hope I didn't annoy you?
Callisto
  • Callisto
Nope, it just the exchange of ideas.
Callisto
  • Callisto
*it's
stamp
  • 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
  • Callisto
The solutions to the system are not numbers, that is it has no unique solution. The link doesn't wok.
Callisto
  • Callisto
*work
mathslover
  • mathslover
@stamp read the question again ;)
anonymous
  • 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
  • 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
  • anonymous
So does the system have infinite solutions or a finite number of solutions?
Callisto
  • Callisto
Infinitely many solution.
anonymous
  • anonymous
If it has infinitely many solutions then which solution gives the smallest value for yz / x^2-3?
anonymous
  • anonymous
I would say range is all real numbers?
Callisto
  • Callisto
No.
anonymous
  • anonymous
no? do you know the answer then?
Callisto
  • Callisto
I know the answer.
Callisto
  • Callisto
I don't know how to get the answer though.
anonymous
  • anonymous
oh ok got you
anonymous
  • 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
  • Callisto
@genius12 Would you mind checking me steps of solving the system?|dw:1363507811134:dw|
Callisto
  • Callisto
*my
anonymous
  • anonymous
Did you do this through gaussian elimination?
Callisto
  • Callisto
Yes.
anonymous
  • 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
  • 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
  • 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
  • Callisto
If it is not infinitely many solutions, then it will be no solution.
Callisto
  • Callisto
You'd better sleep now. Good night.
anonymous
  • anonymous
Then there is infinitely many solutions.
Callisto
  • Callisto
What's next?
anonymous
  • anonymous
Wouldn't the system of equations be inconsistent?
Callisto
  • Callisto
The system should be consistent.
anonymous
  • anonymous
Yes, for you to solve the inequality and get the range for k, it should be consistent which it isn't.
anonymous
  • anonymous
@genius12 yes i agree it is inconsistent.
Callisto
  • Callisto
Please point out my mistakes then.
Callisto
  • Callisto
FYI: http://www.wolframalpha.com/input/?i=x+%2B+2z+%3D1+and+x%2B2y%2B4z+%3D1+and+2x-y%2B3z%3D2
anonymous
  • 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
  • Callisto
That's why we need to think how to get a way to solve it.
anonymous
  • anonymous
Something wrong with the question? lol
Callisto
  • Callisto
No way! It is a past exam question.
anonymous
  • anonymous
What unit does this question cover?
Callisto
  • Callisto
Matrix and determinant.
anonymous
  • anonymous
If it's a past exam, isn't there a given solution?
Callisto
  • Callisto
The final answer - Yes.
anonymous
  • anonymous
Check it? lol
Callisto
  • Callisto
I know what the final answer is. I just don't know how to get the final answer.
Callisto
  • 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
  • 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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