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

• 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
\$\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.
• 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
\$\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.
• 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
\$\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\$
• May 17th 2008, 10:05 PM
mr fantastic
Quote:

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\$

(Bow)