# Roots of quadratic equation lying in a particular range

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 $$m<-b/2a but that didn't helped me to determine the nature of expression $$9a^2 +6ab +4ac$$.

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