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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 \(k(x^2-3) > yz \)
hmm i can find values of x , y and z but don;t know what to do after wards with them
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.
That's what I can do and that's why I ask if you can find the numerical value for {x,y,z}
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.
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?
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.
\[k[(1-2z)^2-3]>-z^2\]\[k>\frac{-z^2}{(1-2z)^2-3}\]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..
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
Oh wait.. Why would I differentiate that!!!
hmm thought u were trying to get max and min.
yeah.. i still feel we using wrong approach.
\[x + 2z =1[1]\] \[x+2y+4z =1[2]\] \[2x-y+3z=2[3]\] 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: \[k(x^2−3)>yz\] \[k(x^2-3)>-\frac{(1-x)^2}{4}\]
\[4k(x^2-3)>-(1-x)^2\] \[4kx^2-12k+(1-2x+x^2)>0\] \[4kx^2-12k+1-2x+x^2>0\] \[(4k+1)x^2-2x-(12k-1)>0\] 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.
Almost there, but still not correct.
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?
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 ] ?
can I post here after 30 minutes please? I have to go for dinner ?
Sure, take your time :)
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?
I am getting diff. answers. Wait!
I have some patience, so I keep waiting, and trying :)
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
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:\[k(x^2-3)>-\frac{(1-x)^2}{4}\]?
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) ?
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 >=
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 ....
lol x=1 btw I was having dinner z=0 and y =-1/2
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
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
So k =-1/6 also satisifies your inequality
Well if z = 0 then go for it: k < 1/2
@aajugdar I don't mind if you check my matrix work above. I can't find a unique solution to the system.
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
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
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 ?
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????
@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)<x or x>(beta) (2) ax^2+bx+c<0, where a>0, then the solution would be like (alpha)<x<(beta)
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
@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)
I am not a man This system has no unique solution, that's what the determinant of coefficient matrix tell us.
well,solve it correctly,it has a solution I found it
@aajugdar You can try using WolframAlpha, you can't get a unique solution. @mathslover Can you guarantee that (4k+1)>0?
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
only in that range x>f(x)
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
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)
Case - 3: 4k+1 = 0 => k = -1/4 So, inequality reduces to z>-1/2
Which obviously consitutes part of real values of z.
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
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
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.
I hope I didn't annoy you?
Nope, it just the exchange of ideas.
it seems the solutions are not numbers, but rather variables (see attachment) also, http://tutorial.math.lamar.edu/Classes/Alg/AugmentedMatrixII.aspx
The solutions to the system are not numbers, that is it has no unique solution. The link doesn't wok.
@stamp read the question again ;)
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
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.
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? do you know the answer then?
I don't know how to get the answer though.
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.
@genius12 Would you mind checking me steps of solving the system?|dw:1363507811134:dw|
Did you do this through gaussian elimination?
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.
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.
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?
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.
Wouldn't the system of equations be inconsistent?
The system should be consistent.
Yes, for you to solve the inequality and get the range for k, it should be consistent which it isn't.
@genius12 yes i agree it is inconsistent.
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
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. =.=
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.
I know what the final answer is. I just don't know how to get the final answer.
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 :(
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.