Polynomials question (1 Viewer)

[unmentionable]

i've been trying to do this question for ages.. its so annoying

the equation x^4 - px^3 + qx^2 - rx +s = 0 has roots alpha, beta, gamma and delta

show that if alpha x beta = gamma x delta, then r^2 = p^2s

thanks

 

[+Po1ntDeXt3r+]

ok conc on r and p and s.. since q isnt req.
alpha=a
beta = b
gamma=y
delta = d

polynomial rule
p = -(alpha + beta + gamma + delta) =-(a+b+y+d)
r = -(aby + byd + yda + dab)
s = abyd

now let ab = yd ("if alpha x beta = gamma x delta")

r^2 = (aby + byd + yda + dab)^2
= (aby + byd)^2 +2(aby + byd)( yda + dab)+ (yda + dab)^2

using the IF thing

=(yd)^2(y+b)^2+2(yd)^2(b+y)(a+d) +(yd)^2(a+d)^2
=(yd)^2 [ (y+b)^2 + 2(b+y)(a+d)+(a+d)^2]
=(yd)^2[ (y+b+a+d)^2]
=(yd)^2[ (y+b+a+d)^2]

s=abyd=(yd)^2 from the ab=yd
and p =a+b+y+d

therefore r^2 = p^2s

 

[+Po1ntDeXt3r+]

just do it systematically..
a question lik this in Q5 last year seemed hard but just think bout the basic steps..
no need to expand it all cos that will make factoring harder.. just think goal is p^2s
not a massive expansion.. massive expansion = bit*h of a factorisation

 

[unmentionable]

thank u so much!

 

[CrashOveride]

or maybe quicker:

abcd = s
a^2.b^2 =s

SIGMA (a) = p
sigma (abc): ab( sigma (a)) = r
little re-arranging, squaring and r^2 = sp^2

 

[CrashOveride]

if a , b and c are roots of x³ + qx + r = 0, find the cubic equation in y, in terms of q and r, whose roots are:

(b+c-2a), (c+a-2b) and (a+b-2c)

Now for these type of questions eg. find eqn whose roots are the squares i would just let y = x² and go from there. But what do i do from here, the solution just uses the first root as y and subs back in and thats it.

 

[CM_Tutor]

if a , b and c are roots of x³ + qx + r = 0, find the cubic equation in y, in terms of q and r, whose roots are:

(b+c-2a), (c+a-2b) and (a+b-2c)

Let y = a + b + c - 3x.

It follows that as x takes the values a, b and c, the roots of the equation in y will be (respectively) b + c - 2a,
c + a - 2b and a + b - 2c.

Now, a + b + c = sum of roots = 0, and so we have y = -3x, or x = -y / 3

(-y / 3)³ + q(-y / 3) + r = 0
On multiplication by -27, we get: y³ + 9qy - 27r = 0, which is the required equation.

 

[+Po1ntDeXt3r+]

or maybe quicker:

abcd = s
a^2.b^2 =s

SIGMA (a) = p
sigma (abc): ab( sigma (a)) = r
little re-arranging, squaring and r^2 = sp^2

how much quicker with sigma form that u propose?

 

[unmentionable]

i did the question using pointdexter's method and it only took me a few minutes
i dont understand the sigma notation thing sorry =\

 

[schmeichung]

Sorry I dont get this:
>>>Let y = a + b + c - 3x.

How would you think of this?

Let y = a + b + c - 3x.

It follows that as x takes the values a, b and c, the roots of the equation in y will be (respectively) b + c - 2a,
c + a - 2b and a + b - 2c.

Now, a + b + c = sum of roots = 0, and so we have y = -3x, or x = -y / 3

(-y / 3)³ + q(-y / 3) + r = 0
On multiplication by -27, we get: y³ + 9qy - 27r = 0, which is the required equation.

 

[+Po1ntDeXt3r+]

he got y is the way to generate the new equations roots.. from the old one ,x.
it took me a while too but im so hungova..

umm lik if i wanted the roots on new eq to be 3a,3b,3c
id let y=3x

 

[CrashOveride]

damn shud have got that, thnx CM

btw, if we have some polynomial P(x) with a zero of multiplicity r, we can express it as P(x) = (x - A)^r . Q(x) ..... where A is the zero. Why do we say also "where Q(A) != 0"

 

[+Po1ntDeXt3r+]

"!=" wats that?

 

[CrashOveride]

!= means doesnt equal to, not equal to

 

[+Po1ntDeXt3r+]

well if Q(A) = 0 then the multiplicity would be >r
i.e. P(x) = (x - A)^r . Q(x)
and Q(A) =0 then A is a factor of Q(x)

i.e.Q(x)= (x-A).M(x) and then P(x)=(x - A)^{r+1} . M(x)
so the multiplicity would not be r.. hope that makes sense

 

[CM_Tutor]

Sorry I dont get this:
>>>Let y = a + b + c - 3x.

How would you think of this?

Most substitutions are fairly obvious.

ie. If you have roots at x = a, b and c, and you want roots at y = a + 1, b + 1 and c + 1, then let y = x + 1

If you have roots at x = a, b and c, and you want roots at y = 5a, 5b and 5c, then let y = 5x

If you have roots at x = a, b and c, and you want roots at y = a², b² and c², then let y = x²

However, some substitutions are more complicated. If the new roots are in terms of more than one existing root, like in the question above, then the trick is that the substitution must be symmetric in the existing roots. That is, you must be able to swap the roots a for b, b for c and c for a (for example) in the substitution expression, whilst the substitution itself remains the same. For example, y = a + b + c - x is symmetric in a, b and c, but y = a + b - c - x is not. Also, if possible, the substiutions should make use of a + b + c, ab + bc + ca or abc, as these are known from the original equation.

In thie question here, I need the roots added together, so I started with y = a + b + c + f(x), where f(x) must be a function of x such that y takes the required values as x takes the roots a, b and c. Now, when x = a, I have
a + b + c + f(a), which needs to be one of b + c - 2a, c + a - 2b or a + b - 2c. Clearly, the only possibility is
a + b + c + f(a) = b + c - 2a, as otherwise f(a) depends on b or c, which isn't possible. So, f(a) = -3a, and the required substitution is y = a + b + c - 3x

Some more examples:

Required Roots: a + b, b + c and c + a. Required Substitution: y = a + b + c - x
Required Roots: 2a + b + c, a + 2b + c and a + 2b + c. Required Substitution: y = a + b + c + x
Required Roots: a²bc, ab²c and abc². Required Substitution: y = abcx
Required Roots: ab, bc and ca. Required Substitution: y = abc / x
Required Roots: ab / c, bc / a and ca / b. Required Substitution: y = abc / x²
Required Roots: a² + b², b² + c² and c² + a². Required Substitution: y = a² + b² + c² - x²
which can be more usefully written as y = (a + b + c)² - 2(ab + bc + ca) - x²

Does this make it clearer?

 

[+Po1ntDeXt3r+]

thats a real good method/ explanation!? i mostly got the obvious ones thru past experiences or it would just pop into my head.. (dun ask.. many eureka moments in 4U last year )..
method is quite systematic too ^^
key to a good tutor

 

[schmeichung]

I got it now thanks a lot!

Most substitutions are fairly obvious.

ie. If you have roots at x = a, b and c, and you want roots at y = a + 1, b + 1 and c + 1, then let y = x + 1

If you have roots at x = a, b and c, and you want roots at y = 5a, 5b and 5c, then let y = 5x

If you have roots at x = a, b and c, and you want roots at y = a², b² and c², then let y = x²

However, some substitutions are more complicated. If the new roots are in terms of more than one existing root, like in the question above, then the trick is that the substitution must be symmetric in the existing roots. That is, you must be able to swap the roots a for b, b for c and c for a (for example) in the substitution expression, whilst the substitution itself remains the same. For example, y = a + b + c - x is symmetric in a, b and c, but y = a + b - c - x is not. Also, if possible, the substiutions should make use of a + b + c, ab + bc + ca or abc, as these are known from the original equation.

In thie question here, I need the roots added together, so I started with y = a + b + c + f(x), where f(x) must be a function of x such that y takes the required values as x takes the roots a, b and c. Now, when x = a, I have
a + b + c + f(a), which needs to be one of b + c - 2a, c + a - 2b or a + b - 2c. Clearly, the only possibility is
a + b + c + f(a) = b + c - 2a, as otherwise f(a) depends on b or c, which isn't possible. So, f(a) = -3a, and the required substitution is y = a + b + c - 3x

Some more examples:

Required Roots: a + b, b + c and c + a. Required Substitution: y = a + b + c - x
Required Roots: 2a + b + c, a + 2b + c and a + 2b + c. Required Substitution: y = a + b + c + x
Required Roots: a²bc, ab²c and abc². Required Substitution: y = abcx
Required Roots: ab, bc and ca. Required Substitution: y = abc / x
Required Roots: ab / c, bc / a and ca / b. Required Substitution: y = abc / x²
Required Roots: a² + b², b² + c² and c² + a². Required Substitution: y = a² + b² + c² - x²
which can be more usefully written as y = (a + b + c)² - 2(ab + bc + ca) - x²

Does this make it clearer?