Quadratic Inequalities (check)
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Find the range of values of k for which x^2 - 2kx - k +6 >0 for all values of k. For me, I obtained -3<k<2 however the actual answeris k<-3 and k >2. But isn't discrimnant <0 since the equation is always postive?
Kind of sure I'm right but i may have faulted somewhere.
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Originally posted by Audi:
Find the range of values of k for which x^2 - 2kx - k +6 >0 for all values of k. For me, I obtained -3<k<2 however the actual answeris k<-3 and k >2. But isn't discrimnant <0 since the equation is always postive?
Kind of sure I'm right but i may have faulted somewhere.
Let me give you a clue about where you went wrong.
As long as the the minimum point of the equation is > 0, the equation will be >0 for all values of x.
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Originally posted by Forbiddensinner:
Let me give you a clue about where you went wrong.
As long as the the minimum point of the equation is > 0, the equation will be >0 for all values of x.
TS already knows that as long as the the minimum point of the equation is > 0, the equation will be >0 for all values of x as TS has said that for the quadratic function to be > 0, the discriminant must be < 0.
Question
Find the range of values of k for which x^2 − 2kx − k +6 >0 for all values of k.
Solution
x^2 – 2kx –k + 6 > 0
x^2 − 2kx + k^2 – k^2 – k + 6 > 0
(x – k)^2 – k^2 – k + 6 > 0
Since (x – k)^2 is always positive (or zero),
– k^2 – k + 6 > 0
(− k + 2) (k + 3) > 0
Since the quadratic function in k > 0and it has a maximum point ie coefficient of k^2 is − 1 , so
− 3 < k < 2
PS : I made a mistake in the earlier answer. The answer is now corrected. Sorry for the mistake.
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Originally posted by Lee012lee:
TS already knows that as long as the the minimum point of the equation is > 0, the equation will be >0 for all values of x as TS has said that for the quadratic function to be > 0, the discriminant must be < 0.
This question is fun and it is not the usual quadratic function and inequality question.
Question
Find the range of values of k for which x^2 − 2kx − k +6 >0 for all values of k.
Solution
x^2 – 2kx –k + 6 > 0
x^2 − 2kx + k^2 – k^2 – k + 6 > 0
(x – k)^2 – k^2 – k + 6 > 0
Since (x – k)^2 is always positive,
– k^2 – k + 6 > 0
(− k + 2) (k + 3) > 0
Since the quadratic function in k > 0,
k > 2 or k < − 3
but then if u sub in value of k which is >2 or <-3 into (− k + 2) (k + 3), it will become, <0 ???
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Originally posted by Forbiddensinner:
Let me give you a clue about where you went wrong.
As long as the the minimum point of the equation is > 0, the equation will be >0 for all values of x.
shi fu dun give answer like this leh, everytime u either give too much or give too little...zzzz....
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Originally posted by MasterMoogle:
but then if u sub in value of k which is >2 or <-3 into (− k + 2) (k + 3), it will become, <0 ???
Sorry, MasterMoogle,I made a mistake.
It should be
(− k + 2) (k + 3) > 0
Since the quadratic function in k > 0 and it has a maximum point ie coefficient of k^2 is − 1 , so
− 3 < k < 2
To check :
Since − 3 < k < 2, let us substitute k = 1 into
x^2 - 2kx - k +6
= x^2 - 2(1)x - 1 +6
= x^2 - 2(1)x + 5
= x^2 - 2(1)x + 1 - 1 + 5
= (x - 1)^2 - 1 + 5
= (x - 1)^2 + 4
Since (x - 1)^2 is always positive (or zero), (x - 1)^2 + 4 > 0
Hence, for x^2 - 2kx - k +6 >0 k will be in the range of − 3 < k < 2.
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Originally posted by Lee012lee:
Sorry, MasterMoogle,I made a mistake.
It should be
(− k + 2) (k + 3) > 0
Since the quadratic function in k > 0 and it has a maximum point ie coefficient of k^2 is − 1 , so
− 3 < k < 2
To check :
Since − 3 < k < 2, let us substitute k = 1 into
x^2 - 2kx - k +6
= x^2 - 2(1)x - 1 +6
= x^2 - 2(1)x + 5
= x^2 - 2(1)x + 1 - 1 + 5
= (x - 1)^2 - 1 + 5
= (x - 1)^2 + 4
Since (x - 1)^2 is always positive (or zero), (x - 1)^2 + 4 > 0
Hence, for x^2 - 2kx - k +6 >0 k will be in the range of − 3 < k < 2.
DON'T APOLOGIZE TO ME!!!!!!!!!!!!!!!
later my shi fu will mar shi wor der
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Originally posted by MasterMoogle:
DON'T APOLOGIZE TO ME!!!!!!!!!!!!!!!
later my shi fu will mar shi wor der
haha, i didn't know forbiddensinner was so fearful de. U scared u loss ur free tutor ah :p
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Originally posted by MasterMoogle:
DON'T APOLOGIZE TO ME!!!!!!!!!!!!!!!
later my shi fu will mar shi wor der
MasterMoogle, no need to worry of no free tutor lah. There are so many free tutors in homework forum lah. They are more qualified and experienced than your sifu lah eg Eagle, Wee_ws, UltimateOnline.
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Originally posted by Seowlah:
MasterMoogle, no need to worry of no free tutor lah. There are so many free tutors in homework forum lah. They are more qualified and experienced than your sifu lah eg Eagle, Wee_ws, UltimateOnline.
Everyone do their bit to teach any student that needs help. I don't see you answering any actual questions at all while forbiddensinner has been very helpful in the few months he has spent here.
Besides spamming long list of tuition agencies and private tutors, i see 0 contribution from you. Your posts are better suited for jobs and employment forum and not homework since this is not the place to advertise for agencies or private tutors.
If you have nothing good to say, just spam somewhere else and not here. If you are not here to ask questions on homework or answer their queries, this forum has no place for you.
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Originally posted by MasterMoogle:
DON'T APOLOGIZE TO ME!!!!!!!!!!!!!!!
later my shi fu will mar shi wor der
I am not that unreasonable.
I am only upset if you mock / made fun of the mistakes of others like a certain someone here on this thread.