least value of $3m+n$












1















If $mx^2+nx+6=0$ does not have two disticnt real



roots. Then least value of $3m+n$ is




Try: Let $3m+n=t,$ Then our equation convert into



$mx^2+(t-3m)x+6=0$



Equation does not have real roots



So we have $(t-3m)^2-24mleq 0$



$$t^2-6mt+9m^2-24mgeq 0$$



now i am struck here, could some help me how to solve it , Thanks






calculus







share|cite|improve this question
























  • Is there something unwritten here disallowing you from taking $m = n = 0$? I'm looking at this and found that $m ge 0$, so $3m + n$ is minimized when $m = n = 0$, which satisfies the "$mx^2 + nx + 6 = 0$ doesn't have two distinct real roots" condition.
    – tilper
    yesterday












  • @tilper $x^2-4x + 6$ also works, giving $3m+n = -1$. So it can be smaller than what you get from $m = n = 0$. That's what's stopping you.
    – Arthur
    yesterday












  • Ah, you're right, $n$ could be negative. Totally spaced on that.
    – tilper
    yesterday
















1















If $mx^2+nx+6=0$ does not have two disticnt real



roots. Then least value of $3m+n$ is




Try: Let $3m+n=t,$ Then our equation convert into



$mx^2+(t-3m)x+6=0$



Equation does not have real roots



So we have $(t-3m)^2-24mleq 0$



$$t^2-6mt+9m^2-24mgeq 0$$



now i am struck here, could some help me how to solve it , Thanks






calculus







share|cite|improve this question
























  • Is there something unwritten here disallowing you from taking $m = n = 0$? I'm looking at this and found that $m ge 0$, so $3m + n$ is minimized when $m = n = 0$, which satisfies the "$mx^2 + nx + 6 = 0$ doesn't have two distinct real roots" condition.
    – tilper
    yesterday












  • @tilper $x^2-4x + 6$ also works, giving $3m+n = -1$. So it can be smaller than what you get from $m = n = 0$. That's what's stopping you.
    – Arthur
    yesterday












  • Ah, you're right, $n$ could be negative. Totally spaced on that.
    – tilper
    yesterday














1












1








1








If $mx^2+nx+6=0$ does not have two disticnt real



roots. Then least value of $3m+n$ is




Try: Let $3m+n=t,$ Then our equation convert into



$mx^2+(t-3m)x+6=0$



Equation does not have real roots



So we have $(t-3m)^2-24mleq 0$



$$t^2-6mt+9m^2-24mgeq 0$$



now i am struck here, could some help me how to solve it , Thanks






calculus







share|cite|improve this question
















If $mx^2+nx+6=0$ does not have two disticnt real



roots. Then least value of $3m+n$ is




Try: Let $3m+n=t,$ Then our equation convert into



$mx^2+(t-3m)x+6=0$



Equation does not have real roots



So we have $(t-3m)^2-24mleq 0$



$$t^2-6mt+9m^2-24mgeq 0$$



now i am struck here, could some help me how to solve it , Thanks






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









D Tiwari

5,3802630




5,3802630












  • Is there something unwritten here disallowing you from taking $m = n = 0$? I'm looking at this and found that $m ge 0$, so $3m + n$ is minimized when $m = n = 0$, which satisfies the "$mx^2 + nx + 6 = 0$ doesn't have two distinct real roots" condition.
    – tilper
    yesterday












  • @tilper $x^2-4x + 6$ also works, giving $3m+n = -1$. So it can be smaller than what you get from $m = n = 0$. That's what's stopping you.
    – Arthur
    yesterday












  • Ah, you're right, $n$ could be negative. Totally spaced on that.
    – tilper
    yesterday


















  • Is there something unwritten here disallowing you from taking $m = n = 0$? I'm looking at this and found that $m ge 0$, so $3m + n$ is minimized when $m = n = 0$, which satisfies the "$mx^2 + nx + 6 = 0$ doesn't have two distinct real roots" condition.
    – tilper
    yesterday












  • @tilper $x^2-4x + 6$ also works, giving $3m+n = -1$. So it can be smaller than what you get from $m = n = 0$. That's what's stopping you.
    – Arthur
    yesterday












  • Ah, you're right, $n$ could be negative. Totally spaced on that.
    – tilper
    yesterday
















Is there something unwritten here disallowing you from taking $m = n = 0$? I'm looking at this and found that $m ge 0$, so $3m + n$ is minimized when $m = n = 0$, which satisfies the "$mx^2 + nx + 6 = 0$ doesn't have two distinct real roots" condition.
– tilper
yesterday






Is there something unwritten here disallowing you from taking $m = n = 0$? I'm looking at this and found that $m ge 0$, so $3m + n$ is minimized when $m = n = 0$, which satisfies the "$mx^2 + nx + 6 = 0$ doesn't have two distinct real roots" condition.
– tilper
yesterday














@tilper $x^2-4x + 6$ also works, giving $3m+n = -1$. So it can be smaller than what you get from $m = n = 0$. That's what's stopping you.
– Arthur
yesterday






@tilper $x^2-4x + 6$ also works, giving $3m+n = -1$. So it can be smaller than what you get from $m = n = 0$. That's what's stopping you.
– Arthur
yesterday














Ah, you're right, $n$ could be negative. Totally spaced on that.
– tilper
yesterday




Ah, you're right, $n$ could be negative. Totally spaced on that.
– tilper
yesterday










1 Answer
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2














We need $n^2le24m$



$3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$






share|cite|improve this answer





















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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    2














    We need $n^2le24m$



    $3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$






    share|cite|improve this answer


























      2














      We need $n^2le24m$



      $3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$






      share|cite|improve this answer
























        2












        2








        2






        We need $n^2le24m$



        $3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$






        share|cite|improve this answer












        We need $n^2le24m$



        $3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        lab bhattacharjee

        223k15156274




        223k15156274






























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