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
Year 13 Maths Help Thread
Announcements  Posted on 


 Follow
 101
 17072016 19:30

 Follow
 102
 17072016 19:36
(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 MobilePost rating:1 
 Follow
 103
 17072016 20:15
(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
Do will it always be graph translations for these type of questions? (If that makes any sense)
Posted from TSR MobileLast edited by kiiten; 17072016 at 20:16. 
 Follow
 104
 17072016 20:26
(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 
 Follow
 105
 17072016 20:31
Posting to watch, since I plan on studying a couple of modules during the Summer.

 Follow
 106
 20072016 15:23
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

 Follow
 107
 20072016 15:29
(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 
 Follow
 108
 20072016 15:56
(Original post by NotNotBatman)
Are you finding the tangent instead of the normal?
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
yy^1 = m(xx^1)
so: y6=0(x4)
which equals y=6 
 Follow
 109
 20072016 16:00
(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
yy^1 = m(xx^1)
so: y6=0(x4)
which equals y=6 
 Follow
 110
 20072016 16:02
(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:ShowAQA:
http://www.aqa.org.uk/subjects/mathematics/alevel
Edexcel:
http://qualifications.pearson.com/en...Exammaterials
OCR:
http://www.ocr.org.uk/qualifications...89278907892/
ExamSolutions:
http://www.examsolutions.net/
PhysicsandMathsTutor SolutionBank:
http://www.physicsandmathstutor.com/...anntextbooks/
For those who become bored by A Level Mathematics and Further Mathematics:
STEP (Sixth Term Examination Paper):
http://www.admissionstestingservice.... ep/aboutstep/
MAT (Mathematics Aptitude Test):
http://www.admissionstestingservice.... mat/aboutmat/
AEA (Advanced Extension Award):
http://qualifications.pearson.com/en...tics2008.html
TeeEm's Special Papers:
http://madasmaths.com/ 
 Follow
 111
 20072016 16:03
(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 
 Follow
 112
 20072016 16:04
(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
yy^1 = m(xx^1)
so: y6=0(x4)
which equals y=6Post rating:1 
 Follow
 113
 20072016 16:11
(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? 
 Follow
 114
 20072016 16:27
(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=4Post rating:1 
 Follow
 115
 20072016 16:32
(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=4Post rating:1 
 Follow
 116
 20072016 16:33
(Original post by NotNotBatman;[url="tel:66512300")
66512300[/url]]You could Just write gradient = 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. 
 Follow
 117
 20072016 16:35
(Original post by NotNotBatman)
You could Just write gradient = 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=4Post rating:1 
 Follow
 118
 25072016 22:08
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
I was thinking of using the general case formula as the order of error is equal to , 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 
 Follow
 119
 28072016 01:01

 Follow
 120
 28072016 02:17
(Original post by Ano123)
Prove that the equation has no integer solutions
Prove further that the equation has only irrational roots..
[Note. No credit is given for using the rational roots theorem.]Spoiler:Show
Therefore
In the event is even; then and are always even. However, is even, and 291 is odd, therefore is always odd. therefore we reach a contradiction.
In the event is odd; then and are always odd. However, is odd, and 291 is odd, therefore is always 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.]
Assume is a rational solution to the above cubic where p,q are real and coprime.
This gives us . 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.Last edited by RDKGames; 28072016 at 03:04. Reason: Proof for the second part, with help from B_9710
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: December 9, 2016
Share this discussion:
Tweet
Related discussions:
 The Current Year 13 Thread 20162017
 The Current Year 13 Thread 20162017
 The Current Year 11 Thread Mark I (20162017)
 The Current Year 11 Thread Mark I (20162017)
 ***Grow Your Grades 2016****
 [23 WEEKS LEFT] 201617 Contract: What Are Your ALevel ...
 [24 WEEKS LEFT] 201617 Contract: What Are Your ALevel ...
 STEP Prep Thread 2017
 STEP Prep Thread 2017
 Doing 2 as levels in year 12 and then the third one in year 13?
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.