least value of $3m+n$
$begingroup$
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
asked Jan 3 at 14:03
DXTDXT
6,0762732
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 14:03
DXTDXT
6,0762732
$endgroup$
$begingroup$
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.
$endgroup$
– tilper
Jan 3 at 14:09
$begingroup$
@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.
$endgroup$
– Arthur
Jan 3 at 14:13
$begingroup$
Ah, you're right, $n$ could be negative. Totally spaced on that.
$endgroup$
– tilper
Jan 3 at 14:14
add a comment |
$begingroup$
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
asked Jan 3 at 14:03
DXTDXT
6,0762732
$endgroup$
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
asked Jan 3 at 14:03
DXTDXT
6,0762732
asked Jan 3 at 14:03
DXTDXT
6,0762732
edited Jan 3 at 14:10
DXT
asked Jan 3 at 14:03
DXTDXT
6,0762732
asked Jan 3 at 14:03
DXTDXT
6,0762732
asked Jan 3 at 14:03
DXTDXT
6,0762732
6,0762732
$begingroup$
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.
$endgroup$
– tilper
Jan 3 at 14:09
$begingroup$
@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.
$endgroup$
– Arthur
Jan 3 at 14:13
$begingroup$
Ah, you're right, $n$ could be negative. Totally spaced on that.
$endgroup$
– tilper
Jan 3 at 14:14
add a comment |
$begingroup$
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.
$endgroup$
– tilper
Jan 3 at 14:09
$begingroup$
@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.
$endgroup$
– Arthur
Jan 3 at 14:13
$begingroup$
Ah, you're right, $n$ could be negative. Totally spaced on that.
$endgroup$
– tilper
Jan 3 at 14:14
$begingroup$
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.
$endgroup$
– tilper
Jan 3 at 14:09
$begingroup$
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.
$endgroup$
– tilper
Jan 3 at 14:09
$begingroup$
@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.
$endgroup$
– Arthur
Jan 3 at 14:13
$begingroup$
@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.
$endgroup$
– Arthur
Jan 3 at 14:13
$begingroup$
Ah, you're right, $n$ could be negative. Totally spaced on that.
$endgroup$
– tilper
Jan 3 at 14:14
$begingroup$
Ah, you're right, $n$ could be negative. Totally spaced on that.
$endgroup$
– tilper
Jan 3 at 14:14
add a comment |
1 Answer
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$begingroup$
We need $n^2le24m$
$3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$
answered Jan 3 at 14:11
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
add a comment |
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1 Answer
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$begingroup$
We need $n^2le24m$
$3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$
answered Jan 3 at 14:11
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
add a comment |
$begingroup$
We need $n^2le24m$
$3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$
answered Jan 3 at 14:11
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
add a comment |
$begingroup$
We need $n^2le24m$
$3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$
answered Jan 3 at 14:11
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
We need $n^2le24m$
$3m+ngedfrac{ n^2}8+ n=dfrac{(n+4)^2-16}8ge-2$
answered Jan 3 at 14:11
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 3 at 14:11
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 3 at 14:11
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 3 at 14:11
lab bhattacharjeelab bhattacharjee
228k15158279
228k15158279
add a comment |
add a comment |
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$begingroup$
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.
$endgroup$
– tilper
Jan 3 at 14:09
$begingroup$
@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.
$endgroup$
– Arthur
Jan 3 at 14:13
$begingroup$
Ah, you're right, $n$ could be negative. Totally spaced on that.
$endgroup$
– tilper
Jan 3 at 14:14