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Math Help - a degree 3 polynomial has two equal roots, prove..

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    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.
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  3. #3
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    Quote Originally Posted by afeasfaerw23231233 View Post
    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
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    Quote Originally Posted by NonCommAlg View Post
    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
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