## Callisto Group Title 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 $$k(x^2-3) > yz$$ one year ago one year ago

1. abb0t

I don't know!! :'(

2. nubeer

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

3. Callisto

Same here...

4. nubeer

lol sorry guess not much help i am

5. Callisto

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

6. nubeer

sure , give me a moment.

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

8. abb0t

Gaussian elimination method.

9. Callisto

|dw:1363361660822:dw|

10. Callisto

|dw:1363361796138:dw|

11. Callisto

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

12. 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.

13. nubeer

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

14. Callisto

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

15. nubeer

|dw:1363362152973:dw|

16. 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.

17. Callisto

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

18. nubeer

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

19. Callisto

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

20. nubeer

can try tht

21. 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

22. Callisto

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

23. nubeer

hmm thought u were trying to get max and min.

24. Callisto

But it won't work?

25. nubeer

yeah.. i still feel we using wrong approach.

26. Azteck

$x + 2z =1[1]$ $x+2y+4z =1[2]$ $2x-y+3z=2[3]$ 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: $k(x^2−3)>yz$ $k(x^2-3)>-\frac{(1-x)^2}{4}$

27. Azteck

$4k(x^2-3)>-(1-x)^2$ $4kx^2-12k+(1-2x+x^2)>0$ $4kx^2-12k+1-2x+x^2>0$ $(4k+1)x^2-2x-(12k-1)>0$ 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.

28. Callisto

Almost there, but still not correct.

29. mathslover

Man at work.

30. AravindG

am I late? :)

31. mathslover

I dont' think so.

32. linshan789

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

33. Callisto

I've tried solving the system above...

34. linshan789

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

35. Callisto

It shouldn't have a unique solution.

36. linshan789

Oh. What level of math is this?

37. Callisto

High school

38. mathslover

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

39. Callisto

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

40. linshan789

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

41. mathslover

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

42. Callisto

No....

43. mathslover

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

44. Callisto

45. mathslover

I am back to work.

46. Callisto

Any guideline?

47. mathslover

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

48. mathslover

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

49. Callisto

No.. diff = different/differentiate?

50. mathslover

different.

51. aajugdar

looks easy wait

52. mathslover

I am getting diff. answers. Wait!

53. Callisto

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

54. mathslover

That is good.

55. shubhamsrg

is it [0, -1/6] ?

56. Callisto

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

57. shubhamsrg

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

58. mathslover

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

59. mathslover

is it wrong^ ?

60. mathslover

@Callisto ?

61. Callisto

It is wrong.

62. shubhamsrg

Did you get what I did ?

63. shubhamsrg

Coz I am not very confident myself! :P

64. Callisto

I didn't get what you did :(

65. Callisto

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

66. 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) ?

67. Callisto

68. mathslover

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

69. Callisto

The inequality itself is > , not >=

70. 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 ....

71. mathslover

@Callisto

72. aajugdar

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

73. aajugdar

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

74. 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

75. mathslover

So k =-1/6 also satisifies your inequality

76. Callisto

What if z=0?

77. mathslover

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

78. aajugdar

yes z=0

79. Callisto

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

80. mathslover

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

81. Callisto

But when z=0, the inequality fails.

82. aajugdar

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

83. aajugdar

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

84. 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 ?

85. aajugdar

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

86. Callisto

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

87. aajugdar

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

88. Callisto

-2 x R1 + R3 -> R3

89. 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)<x or x>(beta) (2) ax^2+bx+c<0, where a>0, then the solution would be like (alpha)<x<(beta)

90. aajugdar

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

91. mathslover

@Callisto very correct

92. aajugdar

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

93. mathslover

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

94. Callisto

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

95. aajugdar

2***

96. aajugdar

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

97. Callisto

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

98. 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

99. aajugdar

-facewall-

100. shubhamsrg

only in that range x>f(x)

101. shubhamsrg

I mean k>f(x)

102. 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

103. 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)

104. mathslover

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

105. mathslover

Which obviously consitutes part of real values of z.

106. 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

107. 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

108. 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.

109. mathslover

110. mathslover

I hope I didn't annoy you?

111. Callisto

Nope, it just the exchange of ideas.

112. Callisto

*it's

113. stamp

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

114. Callisto

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

115. Callisto

*work

116. mathslover

@stamp read the question again ;)

117. genius12

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

118. 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.

119. genius12

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

120. Callisto

Infinitely many solution.

121. genius12

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

122. some_someone

I would say range is all real numbers?

123. Callisto

No.

124. some_someone

no? do you know the answer then?

125. Callisto

126. Callisto

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

127. some_someone

oh ok got you

128. genius12

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.

129. Callisto

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

130. Callisto

*my

131. genius12

Did you do this through gaussian elimination?

132. Callisto

Yes.

133. genius12

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.

134. genius12

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.

135. 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?

136. Callisto

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

137. Callisto

You'd better sleep now. Good night.

138. genius12

Then there is infinitely many solutions.

139. Callisto

What's next?

140. some_someone

Wouldn't the system of equations be inconsistent?

141. Callisto

The system should be consistent.

142. genius12

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

143. some_someone

@genius12 yes i agree it is inconsistent.

144. Callisto

Please point out my mistakes then.

145. Callisto
146. genius12

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. =.=

147. Callisto

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

148. genius12

Something wrong with the question? lol

149. Callisto

No way! It is a past exam question.

150. genius12

What unit does this question cover?

151. Callisto

Matrix and determinant.

152. genius12

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

153. Callisto

154. genius12

Check it? lol

155. Callisto

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

156. 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 :(

157. 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.