Year 13 Maths Help Thread

Announcements Posted on
    Offline

    2
    ReputationRep:
    F(x) =0 the solutions are x=3 and + or - sqrt 2

    For f (x+1)=0 how do you work out the solutions. I know that they are 2 and -1 + or - sqrt 2 but how?

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by kiiten;[url="tel:66467072")
    66467072[/url]]F(x) =0 the solutions are x=3 and + or - sqrt 2

    For f (x+1)=0 how do you work out the solutions. I know that they are 2 and -1 + or - sqrt 2 but how?

    Posted from TSR Mobile
    It's a basic graph translation, so the roots are translated by the same amount in the same direction. X+1 means they are translated by vector [-1,0] and so are the roots
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    It's a basic graph translation, so the roots are translated by the same amount in the same direction. X+1 means they are translated by vector [-1,0] and so are the roots
    Of course!!! I didnt think of that. Thanks
    Do will it always be graph translations for these type of questions? (If that makes any sense)

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by kiiten;[url="tel:66467780")
    66467780[/url]]Of course!!! I didnt think of that. Thanks
    Do will it always be graph translations for these type of questions? (If that makes any sense)

    Posted from TSR Mobile
    They can give you any type of transformation on these except from rotation. But for something like f(x-a)+b it will always be a translation from f(x) by vector [a,b]
    Offline

    2
    ReputationRep:
    Posting to watch, since I plan on studying a couple of modules during the Summer.
    Offline

    3
    ReputationRep:
    I have summer work, can someone help Find the equation of the normal to y=2 (root x) + 8/x at the point (4, 6)I have the answers, and it says x=4 but I keep getting y=6 ????? Thanks
    Offline

    3
    ReputationRep:
    (Original post by Pix2015)
    I have summer work, can someone help Find the equation of the normal to y=2 (root x) + 8/x at the point (4, 6)I have the answers, and it says x=4 but I keep getting y=6 ????? Thanks
    Are you finding the tangent instead of the normal?
    Offline

    3
    ReputationRep:
    (Original post by NotNotBatman)
    Are you finding the tangent instead of the normal?
    No because I get the gradient as 0 and to find the perpendicular 0*-1 = 0

    I got 0 by simplifying the equation to: y=2x^1/2 + 8x^-1

    Then Differentiation: dy/dx = x^-1/2 - 8x^-2

    and then sub 4: dy/dx = 4^-1/2 -8(4)^-2

    so: dy/dx = 0 perpendicular: -1*0 =0
    y-y^1 = m(x-x^1)

    so: y-6=0(x-4)

    which equals y=6
    Offline

    3
    ReputationRep:
    (Original post by Pix2015)
    No because I get the gradient as 0 and to find the perpendicular 0*-1 = 0

    I got 0 by simplifying the equation to: y=2x^1/2 + 8x^-1

    Then Differentiation: dy/dx = x^-1/2 - 8x^-2

    and then sub 4: dy/dx = 4^-1/2 -8(4)^-2

    so: dy/dx = 0 perpendicular: -1*0 =0
    y-y^1 = m(x-x^1)

    so: y-6=0(x-4)

    which equals y=6
    When a line has a gradient of 0, it is parallel to the x axis. The normal to this, which is perpendicular would then be parallel to the y axis, this gradient tends to infinity. So you're right up until you mutliplied the gradient by -1, think about it, the tangent is not going to have the same gradient as the normal is it?
    Offline

    2
    ReputationRep:
    (Original post by Palette)
    This thread is for people who are studying A2 Level Maths and Further Maths and is the successor of the Year 12 Maths Help Thread which was created last year. Although I hope to be relaxing much of the time, I understand that other people want to keep their maths skills refreshed over the summer.

    The rules are the same as the Year 12 Maths Help Thread:

    Rules

    1. Keep the discussion related to maths although I do not mind the occasional physics query.
    2. Please post your full working as this will enable a helper to detect any errors in your method.
    3. If you want to help somebody, don't post full solutions. Your help is meant to act as a springboard, not a featherbed as giving full solutions is unlikely to actually make somebody learn.

    Helpers:
    If you want to be a helper, please state your intention to become one and place a brief reason why.

    List:

    Ano123
    B_9710
    SeanFM
    metrize
    RDKGames
    Zacken

    Relevant Links:
    Spoiler:
    Show
    Might be a bit late, but I'd like to help out where I can . It's fun to help and I have tutored students before in school so I would be able to apply it here .
    Offline

    3
    ReputationRep:
    (Original post by Pix2015)
    I have summer work, can someone help Find the equation of the normal to y=2 (root x) + 8/x at the point (4, 6)I have the answers, and it says x=4 but I keep getting y=6 ????? Thanks
    That's the tangent. As dy/dx=0 at X=4 that means it is a stationary point, a tangent to the stationary point would be y=6 therefore the normal is X=4
    Offline

    3
    ReputationRep:
    (Original post by Pix2015)
    No because I get the gradient as 0 and to find the perpendicular 0*-1 = 0

    I got 0 by simplifying the equation to: y=2x^1/2 + 8x^-1

    Then Differentiation: dy/dx = x^-1/2 - 8x^-2

    and then sub 4: dy/dx = 4^-1/2 -8(4)^-2

    so: dy/dx = 0 perpendicular: -1*0 =0
    y-y^1 = m(x-x^1)

    so: y-6=0(x-4)

    which equals y=6
    If the gradient is A then it's normal is -1/A
    Offline

    3
    ReputationRep:
    (Original post by NotNotBatman)
    When a line has a gradient of 0, it is parallel to the x axis. The normal to this, which is perpendicular would then be parallel to the y axis, this gradient tends to infinity. So you're right up until you mutliplied the gradient by -1, think about it, the tangent is not going to have the same gradient as the normal is it?
    oohhhh yeah I get the concept now.But how would I show that in my working out to finally get the answer x=4
    Offline

    3
    ReputationRep:
    (Original post by Pix2015)
    oohhhh yeah I get the concept now.But how would I show that in my working out to finally get the answer x=4
    You could Just write gradient = \infty so vertical line. equation is x=4, the person marking would assume, you knew what you were doing. For your own benefit drawing a sketch would help, it doesn't matter if it's the actual curve, just the important bits need to help you visualise it.
    Offline

    3
    ReputationRep:
    (Original post by Pix2015)
    oohhhh yeah I get the concept now.But how would I show that in my working out to finally get the answer x=4
    Well I'm unsure how you would show it mathematically due to division by 0 for the normal but I would say that the tangent is completely horizontal thus y=6 therefore the normal is completely vertical and must go through (4,6) therefore X=4
    Offline

    3
    ReputationRep:
    (Original post by NotNotBatman;[url="tel:66512300")
    66512300[/url]]You could Just write gradient = \infty so vertical line. equation is x=4, the person marking would assume, you knew what you were doing. For your own benefit drawing a sketch would help, it doesn't matter if it's the actual curve, just the important bits need to help you visualise it.
    You can't just say the gradient equals infinity, it's not a number that can be used within an equation therefore that would make it mathematically invalid, but the examiner would understand most likely.
    Offline

    3
    ReputationRep:
    (Original post by NotNotBatman)
    You could Just write gradient = \infty so vertical line. equation is x=4, the person marking would assume, you knew what you were doing. For your own benefit drawing a sketch would help, it doesn't matter if it's the actual curve, just the important bits need to help you visualise it.
    (Original post by RDKGames)
    Well I'm unsure how you would show it mathematically due to division by 0 for the normal but I would say that the tangent is completely horizontal thus y=6 therefore the normal is completely vertical and must go through (4,6) therefore X=4
    Ahhh, Thank you, Thankyou
    Offline

    1
    ReputationRep:
    Hi there TSR, I'm not quite sure how to do 2i:
    http://www.mei.org.uk/files/2000papers/nc05ju.pdf
    I'm not quite sure how to show  T_n^*=\frac{1}{3}(4T_{2n}^{ }-T_n)
    I was thinking of using the general case formula 

I=A_{2n}+\frac{1}{2^P-1}-(A_{2n}-A_n)

as the order of error is equal to 2^{P+2}, however can I use this general case formula to show further extrapolation using Romberg's method, increasing the order of error each time, or is the formula only relevant for the first extrapolation?
    Thanks in advance,
    Sal
    Offline

    2
    ReputationRep:
    Prove that the equation  x^3-47x^2+291=0 has no integer solutions.
    Prove further that the equation has only irrational roots..
    [Note. No credit is given for using the rational roots theorem.]
    Offline

    3
    ReputationRep:
    (Original post by Ano123)
    Prove that the equation  x^3-47x^2+291=0 has no integer solutions
    Prove further that the equation has only irrational roots..
    [Note. No credit is given for using the rational roots theorem.]
    I'm not good at proofs of this sort, especially with rationality involved, but I had a good midnight go at it and got some help after losing my original train of thought
    Spoiler:
    Show
    (Original post by Ano123)
    Prove that the equation  x^3-47x^2+291=0 has no integer solutions
    Assume x=\alpha is an integer solution to the above cubic.

    Therefore {\alpha}^3=47{\alpha}^2-291

    In the event \alpha is even; then {\alpha}^3 and {\alpha}^2 are always even. However, 47{\alpha}^2 is even, and 291 is odd, therefore even-odd is always odd. even\not=odd therefore we reach a contradiction.


    In the event \alpha is odd; then {\alpha}^3 and {\alpha}^2 are always odd. However, 47{\alpha}^2 is odd, and 291 is odd, therefore odd-odd is always even. odd\not=even therefore we reach a contradiction yet again.

    Thus we prove there are no integer solutions by means of contradiction. QED.

    (Original post by Ano123)
    Prove further that the equation has only irrational roots..
    [Note. No credit is given for using the rational roots theorem.]
    (Thanks to @B_9710 for bringing my mind back to my lost thought)

    Assume x=\alpha=\frac{p}{q} is a rational solution to the above cubic where p,q are real and coprime.

    This gives us \frac{p^3}{q^3}-47\frac{p^2}{q^2}+291=0 \rightarrow p^3-47p^{2}q+291q^3=0. Since p,q are coprime, the left hand side will always gives an odd number, and since 0 is an even number we arrive at a contradiction therefore all the roots to the cubic are irrational. QED.
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: December 9, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
How are you feeling about doing A-levels?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

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
Study resources

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

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