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# Year 13 Maths Help Thread

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1. 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?

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2. (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?

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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
3. (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)

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4. (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)

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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]
5. Posting to watch, since I plan on studying a couple of modules during the Summer.
6. 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
7. (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?
8. (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
9. (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?
10. (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

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 .
11. (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
12. (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
13. (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
14. (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 = 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.
15. (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
16. (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.
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.
17. (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=4
Ahhh, Thank you, Thankyou
18. 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?
Sal
19. 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.]
20. (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.]
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 has no integer solutions
Assume is an integer solution to the above cubic.

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.]
(Thanks to @B_9710 for bringing my mind back to my lost thought)

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.

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