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# C1 Maths differentiaiton watch

1. Find the two points on the curve Y=2x^3 - 5x^2 +9x -1 where the gradient = 13.

So i differentaited to get 6x^2 -10x +9 I then set that equal to the gradient so 6x^2 -10x +9 =13. Then I turned it into a quadratic by subtracting to get 6x^2 -10x +9 =0 and was going to solve for x that way but its a non calculatro question and the asnweres in thebook are given as a fraction, ut i cant see how to factorise this anyone help?
3. Complete the square
I did intally but doing all those surds and things without a claultor makes it very difficult =/
5. Is it not: 6x^2 - 10x - 4 = 0 (have to subtract the thirteen)

That becomes 3x^2 - 5x - 2 = 0 (dividing by 2)

This can be factorised.
6. (Original post by cpdavis)
Complete the square
Forogt how to do it . I remeber you half the coefficent of b so i would get -5 and you square it all or something
7. (Original post by jaheen22)
Is it not: 6x^2 - 10x - 4 = 0 (have to subtract the thirteen)

That becomes 3x^2 - 5x - 2 = 0 (dividing by 2)

This can be factorised.
:O maybe although im bad at factosriisng it takes me ages =/
8. (Original post by hazbaz)
:O maybe although im bad at factosriisng it takes me ages =/
You need two numbers that multiply together to get -2 and when is one multiplied by 3 they add to give -5.

ANS:
Spoiler:
Show
(3x+1)(x-2)
9. A similar question was posted like this couple of days ago. Why start a new thread on this?
10. 6x^2 -10x +9 =13
6x^2-10x-4=0
3x^2-5x-2=0
(3x+1)(x-2)=0
=]
11. Factorised it to get (-3x -1) (-x +2) so x = 1 or x = -2 so i subsititued these into the y = equation to get y = 5 or y = -39 so the points would be (1,5) or (-2, -39) but in the book it says the answer is (-1/3, -4 17/27) and (2,13)
12. x = 2 and x = -1/3, the x values you got are wrong!
13. (Original post by hazbaz)
I did intally but doing all those surds and things without a claultor makes it very difficult =/
You're doing it wrong.

But if you were to obtain an answer in surd form... that's perfectly fine to leave it like that in the exam.

14. (Original post by boromir9111)
x = 2 and x = -1/3, the x values you got are wrong!
Opp ye it should be (3x +1) (x -2) I understand why x = 2 but not 1/3 woudnt it just be -1 beucase you swap the sign of the +1 ?
15. (Original post by hazbaz)
Opp ye it should be (3x +1) (x -2) I understand why x = 2 but not 1/3 woudnt it just be -1 beucase you swap the sign of the +1 ?
3x+1 = 0
3x = -1
x = -1/3

If you didn't understand that, don't be shy to post here
16. (Original post by boromir9111)
3x+1 = 0
3x = -1
x = -1/3

If you didn't understand that, don't be shy to post here
Ah right ye that makes sense and then i just plug those 2 values in to get Y?
17. (Original post by hazbaz)
Ah right ye that makes sense and then i just plug those 2 values in to get Y?
18. They alos get an asnwer of (2,13)...
19. (Original post by hazbaz)
They alos get an asnwer of (2,13)...
Sub x =2 into the original function and that will give y which is 13 Do you understand how that is done?
20. (Original post by hazbaz)
Find the two points on the curve Y=2x^3 - 5x^2 +9x -1 where the gradient = 13.

So i differentaited to get 6x^2 -10x +9 I then set that equal to the gradient so 6x^2 -10x +9 =13. Then I turned it into a quadratic by subtracting to get 6x^2 -10x +9 =0 and was going to solve for x that way but its a non calculatro question and the asnweres in thebook are given as a fraction, ut i cant see how to factorise this anyone help?
Is this a typo or did you forget to subtract 13 from both sides?

Last equation should read 6x^2 - 10x - 4 = 0
=> 6x^2 - 12x + 2x - 4 = 0
=> 6x(x - 2) + 2(x - 2) = 0
=> (6x + 2)(x - 2) = 0

.: x = 2 or x = -1/3

EDIT:

Oh already been answered. And yea I should have divided through by two but this illustrates the method where you split up the bx term and then factorise it in stages. Which I always found easier lol.

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Updated: September 28, 2010
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