How to graphically depict the possible solutions of a quadratic equation











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I have the following quadratic equation :



$$am^2 + bm + (c_1^2 +c_2^2) =0,$$
where the solution is given by



$$m = frac{-bpmsqrt{b^2-4a(c_1^2+c_2^2)}}{2a}.$$ Here, $Delta>0$.



Thus I have two real roots What I would like to do is to graphically depict possible solutions based on the values of $c_1$ and $c_2$. Here, $c_1$ and $c_2$ are two components of a vector and are real numbers.



Can anyone help me out with this? And is it possible to do that?






roots graphing-functions quadratics







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  • There is a formula for the two (positive) solutions $m_{1,2}$.
    – Wuestenfux
    Nov 28 at 9:26










  • @Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
    – newstudent
    Nov 28 at 9:41










  • Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
    – Wuestenfux
    Nov 28 at 9:42










  • What do you mean by "depict possible solutions"?
    – Todor Markov
    Nov 28 at 9:43










  • @TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
    – newstudent
    Nov 28 at 9:47















up vote
0
down vote

favorite












I have the following quadratic equation :



$$am^2 + bm + (c_1^2 +c_2^2) =0,$$
where the solution is given by



$$m = frac{-bpmsqrt{b^2-4a(c_1^2+c_2^2)}}{2a}.$$ Here, $Delta>0$.



Thus I have two real roots What I would like to do is to graphically depict possible solutions based on the values of $c_1$ and $c_2$. Here, $c_1$ and $c_2$ are two components of a vector and are real numbers.



Can anyone help me out with this? And is it possible to do that?






roots graphing-functions quadratics







share|cite|improve this question
























  • There is a formula for the two (positive) solutions $m_{1,2}$.
    – Wuestenfux
    Nov 28 at 9:26










  • @Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
    – newstudent
    Nov 28 at 9:41










  • Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
    – Wuestenfux
    Nov 28 at 9:42










  • What do you mean by "depict possible solutions"?
    – Todor Markov
    Nov 28 at 9:43










  • @TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
    – newstudent
    Nov 28 at 9:47













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have the following quadratic equation :



$$am^2 + bm + (c_1^2 +c_2^2) =0,$$
where the solution is given by



$$m = frac{-bpmsqrt{b^2-4a(c_1^2+c_2^2)}}{2a}.$$ Here, $Delta>0$.



Thus I have two real roots What I would like to do is to graphically depict possible solutions based on the values of $c_1$ and $c_2$. Here, $c_1$ and $c_2$ are two components of a vector and are real numbers.



Can anyone help me out with this? And is it possible to do that?






roots graphing-functions quadratics







share|cite|improve this question















I have the following quadratic equation :



$$am^2 + bm + (c_1^2 +c_2^2) =0,$$
where the solution is given by



$$m = frac{-bpmsqrt{b^2-4a(c_1^2+c_2^2)}}{2a}.$$ Here, $Delta>0$.



Thus I have two real roots What I would like to do is to graphically depict possible solutions based on the values of $c_1$ and $c_2$. Here, $c_1$ and $c_2$ are two components of a vector and are real numbers.



Can anyone help me out with this? And is it possible to do that?






roots graphing-functions quadratics




roots graphing-functions quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 10:17

























asked Nov 28 at 9:17









newstudent

346




346












  • There is a formula for the two (positive) solutions $m_{1,2}$.
    – Wuestenfux
    Nov 28 at 9:26










  • @Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
    – newstudent
    Nov 28 at 9:41










  • Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
    – Wuestenfux
    Nov 28 at 9:42










  • What do you mean by "depict possible solutions"?
    – Todor Markov
    Nov 28 at 9:43










  • @TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
    – newstudent
    Nov 28 at 9:47


















  • There is a formula for the two (positive) solutions $m_{1,2}$.
    – Wuestenfux
    Nov 28 at 9:26










  • @Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
    – newstudent
    Nov 28 at 9:41










  • Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
    – Wuestenfux
    Nov 28 at 9:42










  • What do you mean by "depict possible solutions"?
    – Todor Markov
    Nov 28 at 9:43










  • @TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
    – newstudent
    Nov 28 at 9:47
















There is a formula for the two (positive) solutions $m_{1,2}$.
– Wuestenfux
Nov 28 at 9:26




There is a formula for the two (positive) solutions $m_{1,2}$.
– Wuestenfux
Nov 28 at 9:26












@Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
– newstudent
Nov 28 at 9:41




@Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
– newstudent
Nov 28 at 9:41












Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
– Wuestenfux
Nov 28 at 9:42




Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
– Wuestenfux
Nov 28 at 9:42












What do you mean by "depict possible solutions"?
– Todor Markov
Nov 28 at 9:43




What do you mean by "depict possible solutions"?
– Todor Markov
Nov 28 at 9:43












@TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
– newstudent
Nov 28 at 9:47




@TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
– newstudent
Nov 28 at 9:47










1 Answer
1






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1
down vote



accepted










Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.



If $Delta gt 0$, then



(1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].



(2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).



Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.



Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.



enter image description here



Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.






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    1 Answer
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    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.



    If $Delta gt 0$, then



    (1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].



    (2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).



    Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.



    Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.



    enter image description here



    Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.



      If $Delta gt 0$, then



      (1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].



      (2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).



      Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.



      Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.



      enter image description here



      Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.



        If $Delta gt 0$, then



        (1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].



        (2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).



        Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.



        Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.



        enter image description here



        Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.






        share|cite|improve this answer














        Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.



        If $Delta gt 0$, then



        (1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].



        (2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).



        Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.



        Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.



        enter image description here



        Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 at 2:34

























        answered Nov 28 at 15:31









        Mick

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        11.7k21641






























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