Quadratic equation_to prove that the roots of equation are rational [on hold]

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If $a, b, c$ are rational, show that the roots of the equation $abc^2x^2 + 3a^2cx + b^2cx – 6a^2 – ab + 2b^2 = 0$, are rational.




quadratics




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put on hold as off-topic by B. Mehta, user21820, Did, Holo, Daniel Fischer♦ 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – B. Mehta, user21820, Did, Holo, Daniel Fischer
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Did you mean your equation as you have written it?
    – Cornman
    Aug 6 at 16:38






  • 6




    Welcome to stackexchange. Hint (I haven't tried). Did you look at the discriminant?
    – Ethan Bolker
    Aug 6 at 16:38










  • "Collect like terms."
    – hardmath
    Aug 6 at 16:39










  • at least give us the roots by solving the equation don't directly ask for an answer
    – James
    Aug 6 at 16:41










  • I've tried with the Discriminant: (3a^2 + b^2)^2c^2 + 4(abc^2)(6a^2+ab-2b^2). But I'm not able to prove that the Discriminant is a perfect square.
    – Gaurav Choudhury
    Aug 6 at 16:42














up vote
-3
down vote

favorite












If $a, b, c$ are rational, show that the roots of the equation $abc^2x^2 + 3a^2cx + b^2cx – 6a^2 – ab + 2b^2 = 0$, are rational.




quadratics




share|cite|improve this question













put on hold as off-topic by B. Mehta, user21820, Did, Holo, Daniel Fischer♦ 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – B. Mehta, user21820, Did, Holo, Daniel Fischer
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Did you mean your equation as you have written it?
    – Cornman
    Aug 6 at 16:38






  • 6




    Welcome to stackexchange. Hint (I haven't tried). Did you look at the discriminant?
    – Ethan Bolker
    Aug 6 at 16:38










  • "Collect like terms."
    – hardmath
    Aug 6 at 16:39










  • at least give us the roots by solving the equation don't directly ask for an answer
    – James
    Aug 6 at 16:41










  • I've tried with the Discriminant: (3a^2 + b^2)^2c^2 + 4(abc^2)(6a^2+ab-2b^2). But I'm not able to prove that the Discriminant is a perfect square.
    – Gaurav Choudhury
    Aug 6 at 16:42












up vote
-3
down vote

favorite









up vote
-3
down vote

favorite











If $a, b, c$ are rational, show that the roots of the equation $abc^2x^2 + 3a^2cx + b^2cx – 6a^2 – ab + 2b^2 = 0$, are rational.




quadratics




share|cite|improve this question













If $a, b, c$ are rational, show that the roots of the equation $abc^2x^2 + 3a^2cx + b^2cx – 6a^2 – ab + 2b^2 = 0$, are rational.





quadratics





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 16:38









Cornman

2,56421128




2,56421128









asked Aug 6 at 16:37









Gaurav Choudhury

12




12




put on hold as off-topic by B. Mehta, user21820, Did, Holo, Daniel Fischer♦ 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – B. Mehta, user21820, Did, Holo, Daniel Fischer
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by B. Mehta, user21820, Did, Holo, Daniel Fischer♦ 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – B. Mehta, user21820, Did, Holo, Daniel Fischer
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Did you mean your equation as you have written it?
    – Cornman
    Aug 6 at 16:38






  • 6




    Welcome to stackexchange. Hint (I haven't tried). Did you look at the discriminant?
    – Ethan Bolker
    Aug 6 at 16:38










  • "Collect like terms."
    – hardmath
    Aug 6 at 16:39










  • at least give us the roots by solving the equation don't directly ask for an answer
    – James
    Aug 6 at 16:41










  • I've tried with the Discriminant: (3a^2 + b^2)^2c^2 + 4(abc^2)(6a^2+ab-2b^2). But I'm not able to prove that the Discriminant is a perfect square.
    – Gaurav Choudhury
    Aug 6 at 16:42
















  • Did you mean your equation as you have written it?
    – Cornman
    Aug 6 at 16:38






  • 6




    Welcome to stackexchange. Hint (I haven't tried). Did you look at the discriminant?
    – Ethan Bolker
    Aug 6 at 16:38










  • "Collect like terms."
    – hardmath
    Aug 6 at 16:39










  • at least give us the roots by solving the equation don't directly ask for an answer
    – James
    Aug 6 at 16:41










  • I've tried with the Discriminant: (3a^2 + b^2)^2c^2 + 4(abc^2)(6a^2+ab-2b^2). But I'm not able to prove that the Discriminant is a perfect square.
    – Gaurav Choudhury
    Aug 6 at 16:42















Did you mean your equation as you have written it?
– Cornman
Aug 6 at 16:38




Did you mean your equation as you have written it?
– Cornman
Aug 6 at 16:38




6




6




Welcome to stackexchange. Hint (I haven't tried). Did you look at the discriminant?
– Ethan Bolker
Aug 6 at 16:38




Welcome to stackexchange. Hint (I haven't tried). Did you look at the discriminant?
– Ethan Bolker
Aug 6 at 16:38












"Collect like terms."
– hardmath
Aug 6 at 16:39




"Collect like terms."
– hardmath
Aug 6 at 16:39












at least give us the roots by solving the equation don't directly ask for an answer
– James
Aug 6 at 16:41




at least give us the roots by solving the equation don't directly ask for an answer
– James
Aug 6 at 16:41












I've tried with the Discriminant: (3a^2 + b^2)^2c^2 + 4(abc^2)(6a^2+ab-2b^2). But I'm not able to prove that the Discriminant is a perfect square.
– Gaurav Choudhury
Aug 6 at 16:42




I've tried with the Discriminant: (3a^2 + b^2)^2c^2 + 4(abc^2)(6a^2+ab-2b^2). But I'm not able to prove that the Discriminant is a perfect square.
– Gaurav Choudhury
Aug 6 at 16:42










1 Answer
1






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up vote
1
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If $abc=0$ it's wrong. Try $c=0$ and $a=b=1$.



But for $abcneq0$ we obtain
$$a^2b^2c^2x^2+abc(3a^2+b^2)x+frac(3a^2+b^2)^24-frac(3a^2+b^2)^24+ab(-6a^2-ab+2b^2)=0$$ or
$$left(abcx+frac3a^2+b^22right)^2-frac(3a^2+4ab-b^2)^24=0.$$
Can you end it now?






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    If $abc=0$ it's wrong. Try $c=0$ and $a=b=1$.



    But for $abcneq0$ we obtain
    $$a^2b^2c^2x^2+abc(3a^2+b^2)x+frac(3a^2+b^2)^24-frac(3a^2+b^2)^24+ab(-6a^2-ab+2b^2)=0$$ or
    $$left(abcx+frac3a^2+b^22right)^2-frac(3a^2+4ab-b^2)^24=0.$$
    Can you end it now?






    share|cite|improve this answer

























      up vote
      1
      down vote













      If $abc=0$ it's wrong. Try $c=0$ and $a=b=1$.



      But for $abcneq0$ we obtain
      $$a^2b^2c^2x^2+abc(3a^2+b^2)x+frac(3a^2+b^2)^24-frac(3a^2+b^2)^24+ab(-6a^2-ab+2b^2)=0$$ or
      $$left(abcx+frac3a^2+b^22right)^2-frac(3a^2+4ab-b^2)^24=0.$$
      Can you end it now?






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        If $abc=0$ it's wrong. Try $c=0$ and $a=b=1$.



        But for $abcneq0$ we obtain
        $$a^2b^2c^2x^2+abc(3a^2+b^2)x+frac(3a^2+b^2)^24-frac(3a^2+b^2)^24+ab(-6a^2-ab+2b^2)=0$$ or
        $$left(abcx+frac3a^2+b^22right)^2-frac(3a^2+4ab-b^2)^24=0.$$
        Can you end it now?






        share|cite|improve this answer













        If $abc=0$ it's wrong. Try $c=0$ and $a=b=1$.



        But for $abcneq0$ we obtain
        $$a^2b^2c^2x^2+abc(3a^2+b^2)x+frac(3a^2+b^2)^24-frac(3a^2+b^2)^24+ab(-6a^2-ab+2b^2)=0$$ or
        $$left(abcx+frac3a^2+b^22right)^2-frac(3a^2+4ab-b^2)^24=0.$$
        Can you end it now?







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 6 at 16:46









        Michael Rozenberg

        88.2k1579180




        88.2k1579180












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