Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    9
    ReputationRep:
    Prove that if y=(x-k)/(x^2-4x-k) can take all values as x varies then 0<k<5.

    Right so I rearranged to form the equation:

    (y)x^2-(4y+1)x+(k-ky)=0
    Then we know for there to be real points on the graph then the discriminant of that equation must be greater than or equal to 0.
    So finding the discriminant gives (-4y-1)^2-4y(k-ky) >0(or equal to)
    Where do I go from here??
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Mathematicus65)
    Prove that if y=(x-k)/(x^2-4x-k) can take all values as x varies then 0<k<5.

    Right so I rearranged to form the equation:

    (y)x^2-(4y+1)x+(k-ky)=0
    Then we know for there to be real points on the graph then the discriminant of that equation must be greater than or equal to 0.
    So finding the discriminant gives (-4y-1)^2-4y(k-ky) >0(or equal to)
    Where do I go from here??
    You need this to be true for all y. So, any thoughts on how you can show a quadratic is always greater than or equal to zero?
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by ghostwalker)
    You need this to be true for all y. So, any thoughts on how you can show a quadratic is always greater than or equal to zero?
    No clue ..
    Offline

    15
    ReputationRep:
    (Original post by Mathematicus65)
    No clue ..
    Here's a clue: can the quadratic be this shaped, roughly speaking? \cap
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by Mpagtches)
    Here's a clue: can the quadratic be this shaped, roughly speaking? \cap
    I don't know how I would determine that???
    • Thread Starter
    Offline

    9
    ReputationRep:
    I put the information into the discriminant to get (16+4k)y^2 + (8-4k)y+1>0

    But don't see how this helps me
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Mathematicus65)
    I put the information into the discriminant to get (16+4k)y^2 + (8-4k)y+1>0

    But don't see how this helps me
    This is your new quadratic, and yes it requires quite a bit of algebra to work out.

    Forget about it for now. In general, how do you show a quadratic is always greater than or equal to zero?

    One method in spoiler
    Spoiler:
    Show

    Complete the square to get:

    a(x+b)^2 +c

    This is always >=0 if c>=o
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by ghostwalker)
    This is your new quadratic, and yes it requires quite a bit of algebra to work out.

    Forget about it for now. In general, how do you show a quadratic is always greater than or equal to zero?

    One method in spoilers

    Spoiler:
    Show

    Complete the square to get:

    a(x+b)^2 +c

    This is always >=0 if c>=o
    Well if I complete the square for the original quadratic in the denominator - (x^2-4x-k) then you get (x-2)^2 - (k+4)..
    Offline

    15
    (Original post by Mathematicus65)
    I put the information into the discriminant to get (16+4k)y^2 + (8-4k)y+1>0

    But don't see how this helps me
    This is the closest you got. You have to calculate a discriminant a second time.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by EricPiphany)
    This is the closest you got. You have to calculate a discriminant a second time.
    Name:  image.jpg
Views: 87
Size:  502.9 KB
    Okay so ive managed to work it out now but my solution is still incorrect due to the Incorrect signs. I have found that k needs to be <= 0 and >=5 not 0<k<5 so what have I done wrong??
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Mathematicus65)
    Well if I complete the square for the original quadratic in the denominator - (x^2-4x-k) then you get (x-2)^2 - (k+4)..
    Not in the original equation, in your new one, involving y and k.

    (Original post by Mathematicus65)
    Name:  image.jpg
Views: 87
Size:  502.9 KB
    Okay so ive managed to work it out now but my solution is still incorrect due to the Incorrect signs. I have found that k needs to be <= 0 and >=5 not 0<k<5 so what have I done wrong??
    You don't want the discriminant to be >=0. This implies that there are roots, and so your inequality, (16+4k)y^2 + (8-4k)y+1>=0 , won't be true for all y.

    You require the discriminant to be <=0, which is why you have the signs the wrong way round.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 17, 2015
Poll
Do you like carrot cake?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.