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Thread: 4U Polynomial question HELP !!

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    Post 4U Polynomial question HELP !!

    If the polynomial x^3+3ax^2+3bx+c=0 has a double root, show that the double root is (c-ab)/(2(a^2-b)) , given that a^2 doesnt equal to b

    Please help Im struggling with this question

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    Junior Member HeroWise's Avatar
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    Re: 4U Polynomial question HELP !!

    I am gonna do it a bit later, a bit busy, but have you tried 3rd derivative getting it to be 0 then testing against second derivative and subbing in the original function?

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    Re: 4U Polynomial question HELP !!

    But since the root has a multiplicity of 2 can it be found in the third derivative?

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    Re: 4U Polynomial question HELP !!

    I read third for some reason im blind

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    Re: 4U Polynomial question HELP !!

    yeah okay can you do it now?

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    Re: 4U Polynomial question HELP !!

    can't do the question right now but i assume that this is how you do it:

    since x^3 + 3ax^2 + 3bx + c = 0 has a double root then this means that p(x) = x^3 + 3ax^2 + 3bx + c has a double zero. let this double zero be y. then p(y) = 0 and p'(y) = 0. this means that y^3 + 3ay^2 + 3by + c = 0 and 3y^2 + 6ay + 3b = 0 -> y^2 + 2ay + b = 0. now we solve this last equation for the double root, y. u will get 2 solutions (since quadratic), then sub each presumed double root into the original equation (p(y)). whichever results in p(y) = 0 is the double root. good luck!

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    Re: 4U Polynomial question HELP !!

    Notice that the required answer includes the constant term . That should be a hint - if we were to differentiate the polynomial, we would lose , so not only would we have to expand the cube of a binomial (which can be really messy), we would also have to rewrite the result to include in order to find the double root, so let's try a different approach.

    Let and let the roots be .

    By the relationships between the roots and co-efficients,







    Rearranging,







    Now:





    Therefore,

    Last edited by fan96; 11 Oct 2018 at 10:25 PM.

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    Re: 4U Polynomial question HELP !!

    While the method above is indeed valid, one could also solve the problem using calculus:

    Let P(x)=x3+3ax2+3bx+c.

    Then P'(x)=3x2+6ax+3b.

    For there to be a double root, there must exist some x such that P(x)=0 and P'(x)=0.

    P'(x)=0 iff 3x2+6ax+3b=0, i.e. x2+2ax+b=0.

    Now, P(x)=0 for the same value of x,

    so x3+3ax2+3bx+c=0,

    i.e. x(x2+2ax+b)+ax2+2bx+c=0.

    x(x2+2ax+b)+a(x2+2ax+b)-2a2x-ab+2bx+c=0.

    But x2+2ax+b=0 (proven above),

    so -2a2x-ab+2bx+c=0,

    i.e. x(2b-2a2)+c-ab=0.

    x=(c-ab)/(2(a2-b)) as a2-b≠0.

    Hence, the double root must be equal to (c-ab)/(2(a2-b)) as required.

    Last edited by BChen65536; 11 Oct 2018 at 10:47 PM.
    r3 and fan96 like this.

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