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Equations of curves and lines question

  • #1
hi, hope you can help with this. For your reference I am 16 doing the nationwide 16 year olds exam (GCSE) so please if you can avoid using stuff I havn't done that'd be great :D

Homework Statement



The x coordinates of the points of intersection of the curve y = x^3 - 2x^2 - 5x and a certain straight line give the solutions to the equation x^3 - 2x^2 - 6x + 1 = 0. Find the equation of the straight line.

Homework Equations



n/a

The Attempt at a Solution



To be honest, I have no idea as to how I'd even approach this question, other than maybe try to first find solutions for x^3 - 2x^2 - 6x + 1 = 0 which I havn't been told how to do. Nor have I been told how to do this type of question either, grrrrrrr. I wish my stupid textbook showed working for the answers...


hope you can help
 

Answers and Replies

  • #2
Hootenanny
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Okay, so you have an unknown straight line which intersects the curve at some unknown point. However, you do know that the straight line must be of the form y=mx + c. Now, if the two curves intersect at some point what can you say about their equations at this point?
 
  • #3
x^3 - 2x^2 - 5x = mx + c ?
 
  • #4
Hootenanny
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x^3 - 2x^2 - 5x = mx + c ?
Correct :approve:, now all that is left to do is solve for m and c.
 
  • #5
sorry but how would i go about solving it? i dunno how to solve when there is a ...^3 in there :S

thnx for the help
 
  • #6
Hootenanny
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No problem, so we have our two equations;

[tex]x^3 - 2x^2 - 5x = mx + c[/tex]

[tex]x^3 - 2x^2 - 5x - mx - c = 0[/tex]

Collecting the coefficients of x;

[tex]x^3 - 2x^2 - (5+m)x - c = 0[/tex]

Now, compare this equation with the one given in the question;
The x coordinates of the points of intersection of the curve y = x^3 - 2x^2 - 5x and a certain straight line give the solutions to the equation x^3 - 2x^2 - 6x + 1 = 0. Find the equation of the straight line.
[tex]x^3 - 2x^2 - (5+m)x - c = 0[/tex]
[tex]x^3 - 2x^2 - 6x +1 = 0[/tex]

Can you go from here?
 
Last edited:
  • #7
HallsofIvy
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No problem, so we have our two equations;

[tex]x^3 - 2x^2 - 5x = mx + c[/tex]

[tex]x^3 - 2x^2 - 5x - mx - c = 0[/tex]

Collecting the coefficients of x;

[tex]x^3 - 3x^2 - (5+m)x - c = 0[/tex]

Now, compare this equation with the one given in the question;


[tex]x^3 - 3x^2 - (5+m)x - c = 0[/tex]
[tex]x^3 - 3x^2 - 6x +1 = 0[/tex]

Can you go from here?
But the one give in the question" was
x3- 2x2- 6x+ 1= 0. not what you give.
 
  • #8
Hootenanny
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Correct as usual Halls, a typo from line 2 to line 3, now duly corrected.
 
  • #9
o rite, i see what you did there, but no unfortunately, i can't do the equations with ...^3, is it factoriseable? cause i guessing it not compelteing the square and it obvious not quadratic formula lol

thnx
 
  • #10
Hootenanny
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Look, for a moment, at the final two lines;

[tex]x^3 - 2x^2 - (5+m)x - c = 0[/tex]
[tex]x^3 - 2x^2 - 6x +1 = 0[/tex]

Can you see that (5+m)=6?
 
  • #11
o crap, misread the question haha soz, yeh i got the answer now

thnx for all the help buddy
 
  • #12
Hootenanny
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  • #13
hey, hope ur not to busy...im new to this site,..cud u guide me here?
 
  • #14
how do i post a ques here- sorry to post here irrelevent of the discussion-but i knida ned help
 
  • #15
Hootenanny
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Hi jay and welcome to the forums,

To post a new topic, select the appropriate forum and click the start new thread button at the top of the forum. If you need any more help getting around you can post in the General Discussion forum :smile:
 
  • #16
HallsofIvy
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On my screen, at least, it is "New Topic", not "start new thread".
 
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