Turn on thread page Beta
 You are Here: Home >< Maths

Solving cubic equations! watch

1. The question is -
Solve the simultaneous equations p + q + r = 6, (p^2) + (q^2) + (r^2) = 26 and pqr = -12.

I was able to make the cubic equation but I don't know how to calculate the values of p, q and r.
2. If you've done it right, the 3 roots of the equation are the values of p, q and r.
3. The equation I got was (x^3) - (6x^2) + (5x) - 12 = 0.

Now I don't know how to calculate its roots? I am supposed to show working.
4. (Original post by Rishabh95)
The question is -
Solve the simultaneous equations p + q + r = 6, (p^2) + (q^2) + (r^2) = 26 and pqr = -12.

I was able to make the cubic equation but I don't know how to calculate the values of p, q and r.
The roots are all integers so you should find it quite easy if you use the factor theorem.
5. (Original post by Rishabh95)
The equation I got was (x^3) - (6x^2) + (5x) - 12 = 0.

Now I don't know how to calculate its roots? I am supposed to show working.
Your equation is slightly incorrect.
6. Ok, I think I got the equation wrong!

I tried another question which is -

Write down the equation whose roots are p, q and r, where p + q + r = 6, qr + rp + pq = 11 and pqr = 6. Hence solve the equation!

The equation I got was (x^3) - (6x^2) + (11x) - 6 = 0 which I verified from the answers.

I am guessing as the question says hence, the answer is straightforward, but I don't know how to calculate!
7. (Original post by Get me off the £\?%!^@ computer)
The roots are all integers so you should find it quite easy if you use the factor theorem.
Do you mean its guesswork?
8. (Original post by Rishabh95)
Do you mean its guesswork?
Well, educated guesswork. But yeah, basically.
9. (Original post by DFranklin)
Well, educated guesswork. But yeah, basically.
So what would you guess in the second question?
10. (Original post by Rishabh95)
So what would you guess in the second question?
Look at the coefficients. I see a 1,-6,11,-6. Does that suggest anything?
11. Is it that - there is one between 1 and -6 and two between -6 and 11?
12. You can find a solution easily enough through intelligent guesswork which is what it sounds like you have done but you're not satisfied with that - is that right?

So you looked at pqr=6 and p+q+r=6 and came up with 1, 2 and 3 - that would seem simple enough - yes?
When the question says "solve" you are being asked to identify the few interesting facts about a curve that may or may not cut either the x or y axis (I'm speaking generally now) so solving means finding where the curve cuts the x-axis, i.e. where y=0. So - if you arrange an ordinary quadratic equation (by way of a simple starter example) such that:-

y = x^2+x-6

and factorise such that:-
y=(x-2)(x+3)

to find where the curve cuts the x-axis you need find the values of x that make y=0. When you have factorised you have:-

y=A x B

and to make y=0 then either A or B must be 0, so with:-
y=(x-2)(x+3)

for instance - either x-2=0 or x+3=0 - solving each of these gives you x=2 or x=-3 and these are the so called roots.

So if the question says the roots are p, q and r and you have intelligently guessed p=1, q=2, r=3 you instantly get the factorized version of your cubic equation:-
y=(x-1)(x-2)(x-3)
which multiplies out to give you:-
(x^3) - (6x^2) + (11x) - 6

Now - is it that you didn't want to rely on an intelligent guess?
You have 3 unknowns and 3 equations so it's a matter of substituting a couple of times to eliminated a couple of variables until you are left with an equation that only has numbers and one variable left, e.g.

rp+qr+pq=11
r(p+q) + pq=11 -- But p+q+r=6 so p+q=6-r
r(6-r) +pq=11 -- and pqr=6 so pq=6/r
r(6-r)+6/r=11 -- nothgin but numbers and 1 variable r.
so you end up having to try and solve
6=r^3 -6r^2 +11r -- factorise it

6 = r (r^2-6r+11)
So again you have 6 = A x B
so A and B could be 1 & 6 or 2 & 3 for instance.
If you look at the quadratic (r^2-6r+11) you can't factorise it as b^2 < 4ac and all this means is that there are no roots - i.e. the x^2 curve does come low enough to intersect with the x-axis. How low does it go? It's easy enough to sketch but a simple differentiation to find the turning point finds the lowest value this quadratic function can have:-

y = r^2-6r+11
dy/dx = 2r-6
So the min value occurs where 2r-6=0 so 2r=6 and r=3.
When you put r=3 in r^2-6r+11 you find the min value this quadratic function can have is 2 (this is B in our 6=AxB)

Remembering we said 6 = r (r^2-6r+11), i.e. 6 = A x B
so A and B could be 1 & 6 or 2 & 3 for instance.
So r must be 3 to get
6 = 3 (2) [where 2 is the min value of our quadratic) from 6 = r (r^2-6r+11)

From then on it's easy to get p and q. I don't know if this is what you were looking for but HTH.
13. x^3 + (alpha+beta+gamma)x^2 + (alphabeta + betaalpha + gammaalpha)x - (alphabetagamma)
14. (Original post by Rishabh95)
Ok, I think I got the equation wrong!

I tried another question which is -

Write down the equation whose roots are p, q and r, where p + q + r = 6, qr + rp + pq = 11 and pqr = 6. Hence solve the equation!

The equation I got was (x^3) - (6x^2) + (11x) - 6 = 0 which I verified from the answers.

I am guessing as the question says hence, the answer is straightforward, but I don't know how to calculate!
x=1 is a solution

(x-1)(x^2 -5x + 6 )

(x-1)(x-2)(x-3) hence x=1,2,3

fairly straightfoward, since you've already done the tricky part finding the equation

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 4, 2011
Today on TSR

Are you living on a tight budget at uni?

From budgets to cutbacks...

University open days

1. University of Cambridge
Wed, 26 Sep '18
2. Norwich University of the Arts
Fri, 28 Sep '18
3. Edge Hill University
Faculty of Health and Social Care Undergraduate
Sat, 29 Sep '18
Poll
Useful resources

Make your revision easier

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

How to use LaTex

Writing equations the easy way

Study habits of A* students

Top tips from students who have already aced their exams

Create your own Study Planner

Never miss a deadline again

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE