polynomial proof question (1 Viewer)

with-chu · May 16, 2010

None of the roots α, β, γ of the equation x^3 + 3px + q = 0 is zero.

Form a monic equation with roots βγ/α, αγ/β and αβ/γ, expressing coefficients in terms of p and q.

I did that, and got x^3 + (p^2)(x^2) - 6px + q = 0
but the answer is different? the coefficient of x^2 is (9p^2)/q instead of p^2.

i subbed x=√(-q/y) into polynomial. What went wrong?

Second part:
Deduce that αβ=γ if and only if (3p - q)^2 + q = 0.

no idea where to begin... sum/product of roots???? I can't make it work out...

Help much appreciated!

with-chu · May 16, 2010

adomad said:
is the question saying that alpha, beta and gamma aren't equal to zero?

Yeah. Non-zero roots.

adomad · May 16, 2010

for the second part, its aksing to find the relation between the co-efficients...

product of roots is alpha X beta X gamma = gamma^2 = -q

hence gamma = sqrt(-q)... then sub it in

with-chu · May 16, 2010

adomad said:
for the second part, its aksing to find the relation between the co-efficients...

product of roots is alpha X beta X gamma = gamma^2 = -q

hence gamma = sqrt(-q)... then sub it in

I don't get it... why is alpha x beta x gamma = gamma^2?

Lukybear · May 16, 2010

cause if ab = y
and aby = -q
then y^2=-q

with-chu · May 16, 2010

can you show me the working please? I still can't get it... the subbing in part

with-chu · May 16, 2010

Another question:

f(x) = x^3 - 3px^2 + 4q

p and q are positive real constants.

show that f(x)=0 has three distinct real roots if and only if p^3 > q

can someone show me the full working please?
________________________________________

I got this now. Nevermind ;P

Lukybear · May 16, 2010

well if y is a root then p(y)=0

with-chu · May 16, 2010

Lukybear said:
well if y is a root then p(y)=0

Yeah I understand that. What I can't get is the algebraic simplification when you sub x=√(-q) into the polynomial...;

The actual problem is that I didn't get the first part right; the polynomial is apparently wrong. But I looked over my working a couple of times and couldn't find an error anywhere so I just went with it. Still wrong

;

adomad · May 16, 2010

ok the first part...
you sub in x= sqrt(-q/y)
$P(x)=x^3 +3px+q$
$$subbing in $x$
$(\frac{-q}{y})^\frac{3}{2} +3p(\frac{-q}{y})^\frac{1}{2}+q=0$
$(\frac{-q}{y})^\frac{3}{2} +3p(\frac{-q}{y})^\frac{1}{2}=-q$

$Upon squaring both sides$
$(\frac{-q}{y})^3 +6p(\frac{-q}{y})^2 +9p^2 (\frac{-q}{y})=q^2$
$\frac{-q^3}{y^3}+ \frac{6pq^2}{y^2} -\frac{9p^2q}{y}=q^2$
$$times by$ $ $ y^3$
$-q^3 +6pq^2y-9p^2qy^2=q^2y^3$
$divide throughout by q^2 $ $ must be monic$
$-q+6py-(\frac{9p^2}{q}) y^2=y^3$

then change the y into x so that it looks nicer

note to self: latex is a biatch

with-chu · May 16, 2010

adomad said:
ok the first part...
you sub in x= sqrt(-q/y)
$P(x)=x^3 +3px+q$
$$subbing in $x$
$(\frac{-q}{y})^\frac{3}{2} +3p(\frac{-q}{y})^\frac{1}{2}+q=0$
$(\frac{-q}{y})^\frac{3}{2} +3p(\frac{-q}{y})^\frac{1}{2}=-q$

$Upon squaring both sides$
$(\frac{-q}{y})^3 +6p(\frac{-q}{y})^2 +9p^2 (\frac{-q}{y})=q^2$
$\frac{-q^3}{y^3}+ \frac{6pq^2}{y^2} -\frac{9p^2q}{y}=q^2$
$$times by$ $ $ y^3$
$-q^3 +6pq^2y-9p^2qy^2=q^2y^3$
$divide throughout by q^2 $ $ must be monic$
$-q+6py-(\frac{9p^2}{q}) y^2=y^3$

then change the y into x so that it looks nicer

ahhh thank you thank you

I realised I made the classic mistake of thinkng my q was a 9 somewhere in my working.

I'll try the 2nd part now

adomad · May 16, 2010

ur welcome... and thanks for the rep

gurmies · May 16, 2010

lyounamu · May 16, 2010

gurmies said:
$\text{Let} \,\, f\left ( x \right )=x^{3}-3px^{2}+4q \\\\ f'\left ( x \right )=3x^{2}-6px \\\\ \text{When} \,\, f'\left ( x \right )=0, \,\, 3x^{2}-6px=0\Leftrightarrow x(x-2p)=0 \\\\ \text{Case One}: \,\, f\left ( 0 \right )> 0; \,\, f\left ( 2p \right )<0 \\\\ 8p^{3}-12p^{3} + 4q<0 \Leftrightarrow p^{3}>q \\\\ \text{Case Two}: \,\, f\left ( 0\right )<0; \,\, f\left ( 2p \right )>0 \\\\ \text{This cannot exist, as it implies that} \,\, q<0 \\\\ \text{Note that I have actually used Bolzano's Theorem here (a special case of the Intermediate Value Theorem)}$

lmao

gurmies · May 16, 2010

lyounamu said:
lmao

Lul

shaon0 · May 17, 2010

gurmies said:
Lul

Taught you well, Thanh Tran has.

polynomial proof question (1 Viewer)

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