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

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

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

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

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

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

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

so, $\displaystyle LHS= 4(b^2-3ac)(a^2-3b) = 4((p^2+2pq)^2-3(2p+q)p^2 q)((2p+q)^2 - 3(p^2+2pq)) =$ ......................
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.

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

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

so, $\displaystyle LHS= 4(b^2-3ac)(a^2-3b) = 4((p^2+2pq)^2-3(2p+q)p^2 q)((2p+q)^2 - 3(p^2+2pq)) =$ ......................
RHS = ...............
so finally LHS=RHS
what is the correct method to do it? thanks.
let $\displaystyle x_0$ be the double root. then $\displaystyle f(x_0)=f'(x_0)=0,$ i.e. $\displaystyle x_0^3+ax_0^2 + bx_0+c=0,$ which we'll call it (1), and

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

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

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

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

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

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

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

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

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

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

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

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

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