$1$ as a common root of a quadratic equation












4












$begingroup$


$$ax^2 + bx + c = 0quadtext{and}quad
bx^2 + cx + a = 0$$
have a common root.



In my book, it says that 1 is a common root for those equation?



Is this correct.



When we plug 1 in both the equations, we get $a+b+c = 0$, it says nothing about 1 being a root. Since we don't know if LHS is zero or not.



Where I'm going wrong here?






quadratics







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
    $endgroup$
    – Minus One-Twelfth
    yesterday












  • $begingroup$
    It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
    $endgroup$
    – Minus One-Twelfth
    yesterday










  • $begingroup$
    Could you clarify if the common root is required to be a real root?
    $endgroup$
    – Servaes
    yesterday
















4












$begingroup$


$$ax^2 + bx + c = 0quadtext{and}quad
bx^2 + cx + a = 0$$
have a common root.



In my book, it says that 1 is a common root for those equation?



Is this correct.



When we plug 1 in both the equations, we get $a+b+c = 0$, it says nothing about 1 being a root. Since we don't know if LHS is zero or not.



Where I'm going wrong here?






quadratics







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
    $endgroup$
    – Minus One-Twelfth
    yesterday












  • $begingroup$
    It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
    $endgroup$
    – Minus One-Twelfth
    yesterday










  • $begingroup$
    Could you clarify if the common root is required to be a real root?
    $endgroup$
    – Servaes
    yesterday














4












4








4


1



$begingroup$


$$ax^2 + bx + c = 0quadtext{and}quad
bx^2 + cx + a = 0$$
have a common root.



In my book, it says that 1 is a common root for those equation?



Is this correct.



When we plug 1 in both the equations, we get $a+b+c = 0$, it says nothing about 1 being a root. Since we don't know if LHS is zero or not.



Where I'm going wrong here?






quadratics







share|cite|improve this question











$endgroup$




$$ax^2 + bx + c = 0quadtext{and}quad
bx^2 + cx + a = 0$$
have a common root.



In my book, it says that 1 is a common root for those equation?



Is this correct.



When we plug 1 in both the equations, we get $a+b+c = 0$, it says nothing about 1 being a root. Since we don't know if LHS is zero or not.



Where I'm going wrong here?






quadratics




quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









TheSimpliFire

13.1k62464




13.1k62464










asked yesterday









user638507user638507

342




342








  • 3




    $begingroup$
    Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
    $endgroup$
    – Minus One-Twelfth
    yesterday












  • $begingroup$
    It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
    $endgroup$
    – Minus One-Twelfth
    yesterday










  • $begingroup$
    Could you clarify if the common root is required to be a real root?
    $endgroup$
    – Servaes
    yesterday














  • 3




    $begingroup$
    Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
    $endgroup$
    – Minus One-Twelfth
    yesterday












  • $begingroup$
    It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
    $endgroup$
    – Minus One-Twelfth
    yesterday










  • $begingroup$
    Could you clarify if the common root is required to be a real root?
    $endgroup$
    – Servaes
    yesterday








3




3




$begingroup$
Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
$endgroup$
– Minus One-Twelfth
yesterday






$begingroup$
Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
$endgroup$
– Minus One-Twelfth
yesterday














$begingroup$
It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
$endgroup$
– Minus One-Twelfth
yesterday




$begingroup$
It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
$endgroup$
– Minus One-Twelfth
yesterday












$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
yesterday




$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
yesterday










3 Answers
3






active

oldest

votes


















4












$begingroup$

From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
    $endgroup$
    – zwim
    yesterday






  • 1




    $begingroup$
    @zwim but then it is not a quadratic
    $endgroup$
    – Maria Mazur
    yesterday



















1












$begingroup$

Let $t$ is the common root,



$at^2+bt+c=0$



$bt^2+ct+a=0$



Solve for $t^2,t$



and use $t^2=(t)^2$ to eliminate $t$



and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    See also : math.stackexchange.com/questions/475354/…
    $endgroup$
    – lab bhattacharjee
    yesterday



















0












$begingroup$

Let $t$ be the common root:
$$at^2+bt+c=0\
bt^2+ct+a=0$$
Subtract and factorize:
$$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
(t-1)(at+a-bt-c)=0 Rightarrow t=1$$






share|cite|improve this answer









$endgroup$














    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
      $endgroup$
      – zwim
      yesterday






    • 1




      $begingroup$
      @zwim but then it is not a quadratic
      $endgroup$
      – Maria Mazur
      yesterday
















    4












    $begingroup$

    From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
      $endgroup$
      – zwim
      yesterday






    • 1




      $begingroup$
      @zwim but then it is not a quadratic
      $endgroup$
      – Maria Mazur
      yesterday














    4












    4








    4





    $begingroup$

    From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.






    share|cite|improve this answer









    $endgroup$



    From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    Maria MazurMaria Mazur

    50k1361124




    50k1361124












    • $begingroup$
      rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
      $endgroup$
      – zwim
      yesterday






    • 1




      $begingroup$
      @zwim but then it is not a quadratic
      $endgroup$
      – Maria Mazur
      yesterday


















    • $begingroup$
      rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
      $endgroup$
      – zwim
      yesterday






    • 1




      $begingroup$
      @zwim but then it is not a quadratic
      $endgroup$
      – Maria Mazur
      yesterday
















    $begingroup$
    rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
    $endgroup$
    – zwim
    yesterday




    $begingroup$
    rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
    $endgroup$
    – zwim
    yesterday




    1




    1




    $begingroup$
    @zwim but then it is not a quadratic
    $endgroup$
    – Maria Mazur
    yesterday




    $begingroup$
    @zwim but then it is not a quadratic
    $endgroup$
    – Maria Mazur
    yesterday











    1












    $begingroup$

    Let $t$ is the common root,



    $at^2+bt+c=0$



    $bt^2+ct+a=0$



    Solve for $t^2,t$



    and use $t^2=(t)^2$ to eliminate $t$



    and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      See also : math.stackexchange.com/questions/475354/…
      $endgroup$
      – lab bhattacharjee
      yesterday
















    1












    $begingroup$

    Let $t$ is the common root,



    $at^2+bt+c=0$



    $bt^2+ct+a=0$



    Solve for $t^2,t$



    and use $t^2=(t)^2$ to eliminate $t$



    and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      See also : math.stackexchange.com/questions/475354/…
      $endgroup$
      – lab bhattacharjee
      yesterday














    1












    1








    1





    $begingroup$

    Let $t$ is the common root,



    $at^2+bt+c=0$



    $bt^2+ct+a=0$



    Solve for $t^2,t$



    and use $t^2=(t)^2$ to eliminate $t$



    and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$






    share|cite|improve this answer









    $endgroup$



    Let $t$ is the common root,



    $at^2+bt+c=0$



    $bt^2+ct+a=0$



    Solve for $t^2,t$



    and use $t^2=(t)^2$ to eliminate $t$



    and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    lab bhattacharjeelab bhattacharjee

    228k15158279




    228k15158279












    • $begingroup$
      See also : math.stackexchange.com/questions/475354/…
      $endgroup$
      – lab bhattacharjee
      yesterday


















    • $begingroup$
      See also : math.stackexchange.com/questions/475354/…
      $endgroup$
      – lab bhattacharjee
      yesterday
















    $begingroup$
    See also : math.stackexchange.com/questions/475354/…
    $endgroup$
    – lab bhattacharjee
    yesterday




    $begingroup$
    See also : math.stackexchange.com/questions/475354/…
    $endgroup$
    – lab bhattacharjee
    yesterday











    0












    $begingroup$

    Let $t$ be the common root:
    $$at^2+bt+c=0\
    bt^2+ct+a=0$$
    Subtract and factorize:
    $$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
    a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
    (t-1)(at+a-bt-c)=0 Rightarrow t=1$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $t$ be the common root:
      $$at^2+bt+c=0\
      bt^2+ct+a=0$$
      Subtract and factorize:
      $$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
      a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
      (t-1)(at+a-bt-c)=0 Rightarrow t=1$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $t$ be the common root:
        $$at^2+bt+c=0\
        bt^2+ct+a=0$$
        Subtract and factorize:
        $$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
        a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
        (t-1)(at+a-bt-c)=0 Rightarrow t=1$$






        share|cite|improve this answer









        $endgroup$



        Let $t$ be the common root:
        $$at^2+bt+c=0\
        bt^2+ct+a=0$$
        Subtract and factorize:
        $$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
        a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
        (t-1)(at+a-bt-c)=0 Rightarrow t=1$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        farruhotafarruhota

        21.8k2842




        21.8k2842






























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