# a degree 3 polynomial has two equal roots, prove..

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• May 17th 2008, 07:13 AM
afeasfaerw23231233
a degree 3 polynomial has two equal roots, prove..
p7 q13
question: a, b, c are real no. prove that if the equation
$x^3+ax^2 +bx+c= 0$
has two equal roots, then $4(b^2-3ac)(a^2-3b)= (9c-ab)^2$

i know my method is not right:
let $f(x) = x^3+ax^2 +bx+c$
= $x^3+(2p+q)x^2+(p^2+2pq)x+p^2q$

so, $LHS= 4(b^2-3ac)(a^2-3b) = 4((p^2+2pq)^2-3(2p+q)p^2 q)((2p+q)^2 - 3(p^2+2pq)) =$ ......................
RHS = ...............
so finally LHS=RHS
what is the correct method to do it? thanks.
• May 17th 2008, 03:30 PM
mr fantastic
Quote:

Originally Posted by afeasfaerw23231233
p7 q13
question: a, b, c are real no. prove that if the equation
$x^3+ax^2 +bx+c= 0$
has two equal roots, then $4(b^2-3ac)(a^2-3b)= (9c-ab)^2$

i know my method is not right:
let $f(x) = x^3+ax^2 +bx+c$
= $x^3+(2p+q)x^2+(p^2+2pq)x+p^2q$

so, $LHS= 4(b^2-3ac)(a^2-3b) = 4((p^2+2pq)^2-3(2p+q)p^2 q)((2p+q)^2 - 3(p^2+2pq)) =$ ......................
RHS = ...............
so finally LHS=RHS
what is the correct method to do it? thanks.

Well, there's always Vieta's equations, substitution and brute force algebra. If I have time I'll try to think of a nicer way.
• May 17th 2008, 09:23 PM
NonCommAlg
Quote:

Originally Posted by afeasfaerw23231233
p7 q13
question: a, b, c are real no. prove that if the equation
$x^3+ax^2 +bx+c= 0$
has two equal roots, then $4(b^2-3ac)(a^2-3b)= (9c-ab)^2$

i know my method is not right:
let $f(x) = x^3+ax^2 +bx+c$
= $x^3+(2p+q)x^2+(p^2+2pq)x+p^2q$

so, $LHS= 4(b^2-3ac)(a^2-3b) = 4((p^2+2pq)^2-3(2p+q)p^2 q)((2p+q)^2 - 3(p^2+2pq)) =$ ......................
RHS = ...............
so finally LHS=RHS
what is the correct method to do it? thanks.

let $x_0$ be the double root. then $f(x_0)=f'(x_0)=0,$ i.e. $x_0^3+ax_0^2 + bx_0+c=0,$ which we'll call it (1), and

$3x_0^2 + 2ax_0+b=0,$ which we'll call it (2). if $x_0=0,$ then (1) and (2) gives us b = c = 0, and what you want

to prove obviously holds. so i'll assume that $x_0 \neq 0.$ multiply (1) by 3 and (2) by $-x_0$ and then add the result

together to get $ax_0^2+2bx_0+3c=0.$ call this (3). now multiply (2) by $a$ and (3) by $-3$ and then add the result

together to get $2(a^2-3b)x_0-(9c-ab)=0.$ call this one (4). now multiply (2) by $3c$ and (3) by $-b$ and then

add the result together to get $(9c-ab)x_0-2(b^2-3ac)=0.$ call this (5). finally multiply (4) by $9c - ab$ and (5)

by $-2(a^2-3b)$ and add the result together to get $4(b^2-3ac)(a^2-3b)=(9c-ab)^2. \ \ \ \square$
• May 17th 2008, 10:05 PM
mr fantastic
Quote:

Originally Posted by NonCommAlg
let $x_0$ be the double root. then $f(x_0)=f'(x_0)=0,$ i.e. $x_0^3+ax_0^2 + bx_0+c=0,$ which we'll call it (1), and

$3x_0^2 + 2ax_0+b=0,$ which we'll call it (2). if $x_0=0,$ then (1) and (2) gives us b = c = 0, and what you want

to prove obviously holds. so i'll assume that $x_0 \neq 0.$ multiply (1) by 3 and (2) by $-x_0$ and then add the result

together to get $ax_0^2+2bx_0+3c=0.$ call this (3). now multiply (2) by $a$ and (3) by $-3$ and then add the result

together to get $2(a^2-3b)x_0-(9c-ab)=0.$ call this one (4). now multiply (2) by $3c$ and (3) by $-b$ and then

add the result together to get $(9c-ab)x_0-2(b^2-3ac)=0.$ call this (5). finally multiply (4) by $9c - ab$ and (5)

by $-2(a^2-3b)$ and add the result together to get $4(b^2-3ac)(a^2-3b)=(9c-ab)^2. \ \ \ \square$

(Bow)