If the roots of $$ax^2+bx+c =0 $$ lie between 1 and 2 then how can if find whether $$9a^2 +6ab +4ac$$ is positive or negative? I tried the problem by using the condition that if roots lie between a certain range $(m,n)$ then $$ af(m)f(n) >0$$ and also $m<-b/2a<n$ but that didn't helped me to determine the nature of expression $$9a^2 +6ab +4ac$$.

  • $\begingroup$ f is an unknown. $\endgroup$ – William Elliot Oct 31 '18 at 2:30
  • $\begingroup$ f() represents a general quadratic function here $\endgroup$ – Nalin Yadav Oct 31 '18 at 5:01
  • $\begingroup$ That's not the part of question.it was part of my attempt $\endgroup$ – Nalin Yadav Oct 31 '18 at 5:02
  • $\begingroup$ Sorry if it confused you $\endgroup$ – Nalin Yadav Oct 31 '18 at 5:02
  • $\begingroup$ Not defining f makes everything from there on useless. $\endgroup$ – William Elliot Oct 31 '18 at 11:12

This claim is false if $a=0$, as the expression is then $0$ and neither positive nor negative. I am assuming in the rest of this proof that $a \neq 0$.

Let $f(x)$ be your original function, $ax^2+bx+c$. We are given that it's roots lie in $[1, 2]$, and therefore there exist $r, s \in [1, 2]$ such that $f(x)=a(x-r)(x-s)=ax^2-2a(s+r)x+asr$. From here, we are then looking to prove that $9a^2+6ab+4ac$ is either always positive or always negative. But $$\begin{aligned} 9a^2+6ab+4ac&=9a^2-12a^2(s+r)+4a^2sr \\ &=a^2(9-12(s+r)+4sr). \end{aligned}$$ Now since $a^2$ is always positive, this is equivalent to asking whether $g(s,r) = 9-12(s+r)+4sr$ is either always positive or always negative on the range $(r,s) \in [1, 2]^2$. Luckily for us, this function is linear in $r$ and $s$, and so it suffices to check the corners: $$\begin{aligned} g(1, 1)&=-11 \\ g(1, 2)&=-19 \\ g(2, 1)&=-19 \\ g(2, 2)&=-23 \end{aligned}$$

and so $g$ is always negative

  • $\begingroup$ @ DreamConspiracy "This claim is false if a=0, as the expression is then 0 and neither positive nor negative " This the most elementary assumption we make before studying quadratic equation that leading coefficient is not zero . I didn't wrote it as I thought it was obvious $\endgroup$ – Nalin Yadav Nov 1 '18 at 4:38
  • 1
    $\begingroup$ @NalinYadav good point. I'll keep it in there for clarity anyway $\endgroup$ – DreamConspiracy Nov 1 '18 at 4:51

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