# Equations of curves and lines question

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

## 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.

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

Staff Emeritus
Gold Member
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?

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

Staff Emeritus
Gold Member
x^3 - 2x^2 - 5x = mx + c ?
Correct , now all that is left to do is solve for m and c.

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

Staff Emeritus
Gold Member
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;
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.

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

Can you go from here?

Last edited:
Homework Helper
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;

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

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

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

Can you go from here?

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

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

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

Staff Emeritus
Gold Member
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?

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

thnx for all the help buddy

Staff Emeritus
Gold Member
thnx for all the help buddy pleasure

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

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

Staff Emeritus
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 