Question

If the roots of the equation
\[\large\rm x^4-px^3+qx^2-pqx+1=0\]
are a,b,c and d, show that
\[\large\rm (a+b+c)(a+b+d)(b+c+d)(a+c+d)=1\]

Answer 1

I dont know if this approach is correct one but this is my view.
Let x = 1 (since the equation is true for every value of x) which gives us
1-p+q-pq+1=0
p-q+pq=2
Where p = Σa q= Σab
And also pq=Σabc (I dont think this equation would help)

Answer 2

The given equation is not true for every value of x. It is only true for four values of x.

Answer 3

namely a, b, c and d.

Answer 4

right

Answer 5

Step1 :
Notice that a + b + c + d = p.

Therefore
(a+b+c)(a+b+d)(b+c+d)(a+c+d) = (p-a)(p-b)(p-c)(p-d)

Answer 6

@ganeshie8 We have to show the resulting equation. Thus our solution should start from given the equation. Not the way backwards.

Answer 7

ganeshie8 but we know that than a,b,c and d are roots of this equation so this mean that
(x-a)(x-b)(x-c)(x-d)= x^4 -px^3 +qx^2 -pqx +1 too

yes ?

Answer 8

I took only the left hand side. Not the entire equation.

Answer 9

@jhonyy9 yes. But here we have too many unnecessary variables

Answer 10

@ganeshie8 Okay

Answer 11

Do you still feel the starting of the proof is wrong ?

Answer 12

@ganeshie8 possible one bad idea but dont help here using the Viéte formulas ?

Answer 13

No its just that, we are not allowed to take in account with the given solution. As if we never heard of given solution

Answer 14

Yeah, I'm also trying to use Viete formulas and see if we can simplify

Answer 15

ok. so this mean that make sense really ?

Answer 16

Are you saying we cannot start the proof with left hand side of the statement that we want to prove ?

Answer 17

Yes

Answer 18

Why not ?

Answer 19

Do you think it is wrong to do that or you just don't want to do it that way ?

Answer 20

young man this is not the place to cheat on your test

Answer 21

@ganeshie8 Btw I only care about solution. I dont care if it came by bribing a mathematician so you should carry on with your solution. My opinion:- It feels okay. If we prove right to be equal to left it pretty much means the same

Answer 22

@DrMoss Not a test. It just I have bounded myself with some rules so I wont be surprised if a type of question came up in which this tactic rendered unusable.

Answer 23

Okay, so we're good with step1. Lets keep going.

Answer 24

OKay

Answer 25

Step2 :
(p-a)(p-b)(p-c)(p-d) is a polynomial in "p" whose roots are a, b, c, d.
So we can get its expansion simply by replacing "x" by "p" in the given polynomial :

(p-a)(p-b)(p-c)(p-d) = p^4-p(p)^3+q(p)^2-pq(p)+1

Answer 26

simplify and you're done!

Answer 27

By simplification the left side is
\[\large\rm p^4-(Σa) p^3+(Σab) p^2-(Σabc)p+abcd\ \]

Answer 28

I suggest you go through steps 1 and 2 again.

Answer 29

Which side are we suppose to simplify? Because I dont think any side needs simplification

Answer 30

(p-a)(p-b)(p-c)(p-d) = p^4-p(p)^3+q(p)^2-pq(p)+1 = ?

Answer 31

@FaiqRaees
It would be nice if we are given the exact and complete wording of the question.
A lot of energy would be wasted if people suggest a solution to be told that it is "not allowed". Math is all about precision, so please give all the parameters of the question to have a level playing field.

Answer 32

@ganeshie8 How is p=a+b+c+d = (p-a)(p-b)(p-c)(p-d)?

Answer 33

I never said p = (p-a)(p-b)(p-c)(p-d)

Answer 34

a + b + c + d = p

gives you

a + b + c = p - d
b + c + d = p - a
c + d + a = p - b
d + a + b = p - c

Answer 35

@mathmate The question wordings are that exactly. It was just that my teacher didn't accepted solution in which instead of simplifying a given equation, the proof was given by expanding the given solution

Answer 36

Oh okay that clears a lot

Answer 37

So now if I can show that
(p-a)(p-b)(p-c)(p-d) = p^4-p(p)^3+q(p)^2-pq(p)+1
my work is done right?

Answer 38

Ohkay, I see... I should have added more steps

Answer 39

I have already shown that for you in step2.

Answer 40

Let me explain step2 again.

Answer 41

Yeah that would be very helpful

Answer 42

We're given that the roots of the polynomial x^4-px^3+qx^2-pqx+1 are a, b, c, d.

So we have
x^4-px^3+qx^2-pqx+1 = (x-a)(x-b)(x-c)(x-d)

Answer 43

With me ?

Answer 44

Oh got it

Answer 45

Replace x with p

Answer 46

Yep, simple once you see it :)

Answer 47

And since we have already proven that those p brackets are just a simplified version of the given solution, the equation equals the solution

Answer 48

But our proof isn't complete yet I suppose. We have to prove that the given solution is equal to 1. We haven't done that or did we?

Answer 49

(p-a)(p-b)(p-c)(p-d) = p^4-p(p)^3+q(p)^2-pq(p)+1 = ?

Answer 50

0

Answer 51

seriously you should have some rest -.-

Answer 52

That is given in the equation
x^4-px^3+qx^2-pqx+1=0

Answer 53

Again, that equation is NOT an identity. It is not true for every value of x. It is true only when x = a, b, c, d.

Answer 54

Oh yeah, right, I keep forgetting that. But if we input p, then it should output 1. Yeah okay understood thanks

Answer 55

Why ?

Answer 56

Algebraic manipulation(I loathe myself for missing such thing)

Answer 57

Happens haha. I'm just making sure you got it right :)

Answer 58

@ganeshie8 Yeah thanks

Answer 59

(x-a)(x-b)(x-c)(x-d)= x^4 -px^3 +qx^2 -pqx +1 = x^3(x-p)+qx(x-p) +1 =
= (x^3 +qx)(x-p) +1

this dont help anything ?

Answer 60

@ganeshie8 I have a confusion for which the book is giving different answer and the people at math.stackexchange are giving different answer. Can I confirm it with you if you have time?

Answer 61

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
(x-a)(x-b)(x-c)(x-d)= x^4 -px^3 +qx^2 -pqx +1 = x^3(x-p)+qx(x-p) +1 =
= (x^3 +qx)(x-p) +1

this dont help anything ?
\(\color{#0cbb34}{\text{End of Quote}}\)
I suppose unless there is an x in the expression, it cant help anything. That's maybe why @ganeshie8 first took steps to eliminate x out of the equations

Answer 62

@FaiqRaees are you on stackexchange too ?

Answer 63

I'll try, ask

Answer 64

@jhonyy9 Yeah because the good members here come online very late at night(according to my timezone)

Answer 65

how ? are you from Europe same with me ?

Answer 66

@ganeshie8 What is the general formula for the coefficient of x^n-2 of a polynomial f(x) having the degree of n?

Answer 67

@jhonyy9 Asia

Answer 68

ok i understand now

Answer 69

Do you mean the coefficient of x^(n-2) ?

Answer 70

Yes

Answer 71

What things do we know ?

Answer 72

Are we given the roots of the polynomial ?

Answer 73

No I am just asking generally like in the form Σab or in that way

Answer 74

You must ask the question clearly. Then only we can pursue the answer...

Answer 75

If we knew the roots, we could simply use the Vieta formulas to find the coefficients.

Answer 76

I dont really know how to describe it
Like for an equation
x^n +ax^n-1+bx^n-2......
The coefficient of x^(n-1)is Σa
What is for x^(n-2)?

Answer 77

Okay, I see what you're asking.

Answer 78

Consider the polynomial
\[P(x) = x^n + a_{n-1} x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1 x^1 + a_0 \]
Let \(\beta_1, \beta_2, \ldots , \beta_n\) be the roots of \(P(x)\).

Answer 79

Okay

Answer 80

Then we can factor P(x) as
\[x^n + a_{n-1} x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1 x^1 + a_0 = (x-\beta_1)(x - \beta_2)\cdots (x-\beta_n) \]

Answer 81

True

Answer 82

What is the coefficient of x^n on left hand side ?

Answer 83

1

Answer 84

What is the coefficient of x^n on right hand side ?

Answer 85

1

Answer 86

They are equal. So we're good with the coefficients of x^n.
Let's compare the coefficients of x^(n-1) both sides.

Answer 87

What is the coefficient of x^(n-1) on left hand side ?

Answer 88

a(n-1)

Answer 89

What is the coefficient of x^(n-1) on right hand side ?

Answer 90

it should be a(n-1) (considering we are allowed to assume the equality to be true otherwise no way to find out, at least for me)

Answer 91

Right hand side, you should get an expression in terms of the roots. Lets see

Answer 92

Our goal is to find the coefficient of x^(n-1) in
\[ (x-\beta_1)(x - \beta_2)\cdots (x-\beta_n) \]

Answer 93

How many factors are there in that product ?

Answer 94

n

Answer 95

Yes. How did we find the coefficient of x^n ?

Answer 96

For right side? By simple logic. There is no coefficient with x in any of the factors. To reach x^n, the x terms can only multiply with x terms. Since there is no coefficient on any value of x thus the coefficient of x^n is 1.
Or you can also say that by just comparing it with left side