# Equations of curves and lines question

1. Feb 25, 2007

### Trail_Builder

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

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

n/a

3. 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

2. Feb 25, 2007

### Hootenanny

Staff Emeritus
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. Feb 25, 2007

### Trail_Builder

x^3 - 2x^2 - 5x = mx + c ?

4. Feb 25, 2007

### Hootenanny

Staff Emeritus
Correct , now all that is left to do is solve for m and c.

5. Feb 25, 2007

### Trail_Builder

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. Feb 25, 2007

### Hootenanny

Staff Emeritus
No problem, so we have our two equations;

$$x^3 - 2x^2 - 5x = mx + c$$

$$x^3 - 2x^2 - 5x - mx - c = 0$$

Collecting the coefficients of x;

$$x^3 - 2x^2 - (5+m)x - c = 0$$

Now, compare this equation with the one given in the question;
$$x^3 - 2x^2 - (5+m)x - c = 0$$
$$x^3 - 2x^2 - 6x +1 = 0$$

Can you go from here?

Last edited: Feb 25, 2007
7. Feb 25, 2007

### HallsofIvy

Staff Emeritus
But the one give in the question" was
x3- 2x2- 6x+ 1= 0. not what you give.

8. Feb 25, 2007

### Hootenanny

Staff Emeritus
Correct as usual Halls, a typo from line 2 to line 3, now duly corrected.

9. Feb 25, 2007

### Trail_Builder

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. Feb 25, 2007

### Hootenanny

Staff Emeritus
Look, for a moment, at the final two lines;

$$x^3 - 2x^2 - (5+m)x - c = 0$$
$$x^3 - 2x^2 - 6x +1 = 0$$

Can you see that (5+m)=6?

11. Feb 25, 2007

### Trail_Builder

o crap, misread the question haha soz, yeh i got the answer now

thnx for all the help buddy

12. Feb 25, 2007

### Hootenanny

Staff Emeritus
pleasure

13. Feb 25, 2007

### jay ambekar

hey, hope ur not to busy...im new to this site,..cud u guide me here?

14. Feb 25, 2007

### jay ambekar

how do i post a ques here- sorry to post here irrelevent of the discussion-but i knida ned help

15. Feb 25, 2007

### Hootenanny

Staff Emeritus
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

16. Feb 25, 2007

### HallsofIvy

Staff Emeritus
On my screen, at least, it is "New Topic", not "start new thread".