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If $ax^2+bx+c=0$ has no real roots then find min value of $3a+b$ for $c=6$;

Please tell me how to proceed , i don't have any clue of what to do.

ABC
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  • From a language point of view, there is no minimum, only a greatest lower bound. That is, we can get arbitrarily close to the lower bound, but can't actually reach it, I believe. – Thomas Andrews Mar 25 '13 at 16:14

3 Answers3

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So, we need the discriminant $b^2-4\cdot a\cdot6<0\implies b^2<24a$

So, $24a>b^2, 24a=b^2+\epsilon$ where real number $\epsilon>0$

So, $8(3a+b)=24a+8b=\epsilon+b^2+8b=\epsilon+(b+4)^2-16\ge \epsilon -16$

So, $3a+b>\frac{\epsilon-16}8=\frac{\epsilon}8-2$

Now, $\epsilon$ can be made arbitrarily small keeping it positive.

So, $3a+b$ can be made arbitrarily close to $-2,$ but always $>-2$

So, there is no minimum value as such unless further restriction(s) (like $a,b$ are integer etc.) are included.

lab bhattacharjee
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    that only shows that $-2$ is a lower bound, but it isn't much harder to show it is a greatest lower bound. – Thomas Andrews Mar 25 '13 at 16:18
  • So it was just some manipulations' play ...... – ABC Mar 25 '13 at 16:19
  • @ThomasAndrews, $3a+b$ can not be $=-2$ right? It just can be set to closest to $-2$ – lab bhattacharjee Mar 25 '13 at 16:21
  • Right, my point was, you didn't show that it can be arbitrarily close to $-2$, you just showed it had a lower bound at $-2$. You could show that $-3$ is a lower bound as well. To show that $-2$ is the greatest lower bound, you need to show that, for any $\epsilon>0$ you can find such $a,b$ with $3a+b<-2+\epsilon$. @labbhattacharjee – Thomas Andrews Mar 25 '13 at 16:33
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The equation has non-real roots whenever $b^2 < 4ac$. Since you are given $c=6$, the problem is $\inf \{ 3a+b | b^2 < 24 a \}$.

The problem has no minimum, because if $(a,b)$ solved the problem, then since $b^2<24a$, we can reduce $a$ by a small amount and produce a smaller value. So we 'relax' the problem to $\inf \{ 3a+b | b^2 \le 24 a \}$.

Notice, as above, that if $b^2<24 a$, then $a$ can be changed slightly to reduce the objective. Hence either the problem is unbounded below, or the constraint must be active. However, if $b^2<24 a$, then $3a+b > \frac{1}{8}b^2+b$, and the latter is bounded below because it is a quadratic with positive leading coefficient.

So, the problem is equivalent to $\inf \{ 3a+b | b^2 = 24 a \}= \inf \{ \frac{1}{8}b^2+b | b^2 = 24 a \}=\inf \frac{1}{8}b^2+b = -2$.

copper.hat
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This question focuses on the discriminant of a quadratic function. The discriminant, let's call it $d$, is defined as $b^2 - 4ac$. In this case, there are no real roots. Hence, the discriminant is less than $0$, or $d<0$.

Considering that $c$ = 6, you can form an inequality $$ \begin{align*} b^2 - 4ac &<0\\ b^2 - 24a &<0\\ \end{align*} $$

Find the minimum of $3a+b$ which satisfies the above inequality.

fahrbach
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bryan.blackbee
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