Finding the minimum value.












3












$begingroup$


I'm struck on this question, I tried hard but couldn't solve it.



Question: if a quadratic equation in $x$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $5a + b$.



So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.



Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $x.$



Any help will be appreciated.






calculus algebra-precalculus quadratics quadratic-forms







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I'm struck on this question, I tried hard but couldn't solve it.



    Question: if a quadratic equation in $x$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $5a + b$.



    So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.



    Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $x.$



    Any help will be appreciated.






    calculus algebra-precalculus quadratics quadratic-forms







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      0



      $begingroup$


      I'm struck on this question, I tried hard but couldn't solve it.



      Question: if a quadratic equation in $x$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $5a + b$.



      So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.



      Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $x.$



      Any help will be appreciated.






      calculus algebra-precalculus quadratics quadratic-forms







      share|cite|improve this question











      $endgroup$




      I'm struck on this question, I tried hard but couldn't solve it.



      Question: if a quadratic equation in $x$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $5a + b$.



      So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.



      Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $x.$



      Any help will be appreciated.






      calculus algebra-precalculus quadratics quadratic-forms




      calculus algebra-precalculus quadratics quadratic-forms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 12:22









      amWhy

      1




      1










      asked Dec 8 '18 at 12:15









      Shivansh JShivansh J

      346




      346






















          1 Answer
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          7












          $begingroup$

          $$Dleq0$$
          $$Longrightarrow b^2leq20a$$
          $$Longrightarrow 5ageqfrac{b^2}{4}$$
          $$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
          where equality holds if $b = -2 and a = frac{1}{5}$



          Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.



          Hope it is helpful:)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
            $endgroup$
            – s0ulr3aper07
            Dec 8 '18 at 12:59






          • 1




            $begingroup$
            Yes, that is how I did it.
            $endgroup$
            – Martund
            Dec 8 '18 at 13:01










          • $begingroup$
            Thanks for help..finally I know how to deal with these problems.
            $endgroup$
            – Shivansh J
            Dec 8 '18 at 16:03











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          $begingroup$

          $$Dleq0$$
          $$Longrightarrow b^2leq20a$$
          $$Longrightarrow 5ageqfrac{b^2}{4}$$
          $$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
          where equality holds if $b = -2 and a = frac{1}{5}$



          Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.



          Hope it is helpful:)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
            $endgroup$
            – s0ulr3aper07
            Dec 8 '18 at 12:59






          • 1




            $begingroup$
            Yes, that is how I did it.
            $endgroup$
            – Martund
            Dec 8 '18 at 13:01










          • $begingroup$
            Thanks for help..finally I know how to deal with these problems.
            $endgroup$
            – Shivansh J
            Dec 8 '18 at 16:03
















          7












          $begingroup$

          $$Dleq0$$
          $$Longrightarrow b^2leq20a$$
          $$Longrightarrow 5ageqfrac{b^2}{4}$$
          $$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
          where equality holds if $b = -2 and a = frac{1}{5}$



          Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.



          Hope it is helpful:)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
            $endgroup$
            – s0ulr3aper07
            Dec 8 '18 at 12:59






          • 1




            $begingroup$
            Yes, that is how I did it.
            $endgroup$
            – Martund
            Dec 8 '18 at 13:01










          • $begingroup$
            Thanks for help..finally I know how to deal with these problems.
            $endgroup$
            – Shivansh J
            Dec 8 '18 at 16:03














          7












          7








          7





          $begingroup$

          $$Dleq0$$
          $$Longrightarrow b^2leq20a$$
          $$Longrightarrow 5ageqfrac{b^2}{4}$$
          $$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
          where equality holds if $b = -2 and a = frac{1}{5}$



          Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.



          Hope it is helpful:)






          share|cite|improve this answer











          $endgroup$



          $$Dleq0$$
          $$Longrightarrow b^2leq20a$$
          $$Longrightarrow 5ageqfrac{b^2}{4}$$
          $$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
          where equality holds if $b = -2 and a = frac{1}{5}$



          Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.



          Hope it is helpful:)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 13:50

























          answered Dec 8 '18 at 12:49









          MartundMartund

          1,453212




          1,453212








          • 1




            $begingroup$
            Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
            $endgroup$
            – s0ulr3aper07
            Dec 8 '18 at 12:59






          • 1




            $begingroup$
            Yes, that is how I did it.
            $endgroup$
            – Martund
            Dec 8 '18 at 13:01










          • $begingroup$
            Thanks for help..finally I know how to deal with these problems.
            $endgroup$
            – Shivansh J
            Dec 8 '18 at 16:03














          • 1




            $begingroup$
            Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
            $endgroup$
            – s0ulr3aper07
            Dec 8 '18 at 12:59






          • 1




            $begingroup$
            Yes, that is how I did it.
            $endgroup$
            – Martund
            Dec 8 '18 at 13:01










          • $begingroup$
            Thanks for help..finally I know how to deal with these problems.
            $endgroup$
            – Shivansh J
            Dec 8 '18 at 16:03








          1




          1




          $begingroup$
          Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
          $endgroup$
          – s0ulr3aper07
          Dec 8 '18 at 12:59




          $begingroup$
          Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
          $endgroup$
          – s0ulr3aper07
          Dec 8 '18 at 12:59




          1




          1




          $begingroup$
          Yes, that is how I did it.
          $endgroup$
          – Martund
          Dec 8 '18 at 13:01




          $begingroup$
          Yes, that is how I did it.
          $endgroup$
          – Martund
          Dec 8 '18 at 13:01












          $begingroup$
          Thanks for help..finally I know how to deal with these problems.
          $endgroup$
          – Shivansh J
          Dec 8 '18 at 16:03




          $begingroup$
          Thanks for help..finally I know how to deal with these problems.
          $endgroup$
          – Shivansh J
          Dec 8 '18 at 16:03


















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