If $a$ and $4a+3b+2c$ have same sign, then which given interval can not contain roots of $ax^2+bx+c=0$?

Question

If $a$ and $4a+3b+2c$ have same sign, then $$ax^2+bx+c=0$$ (with $a\neq0$) can not have roots belonging to:

(a) $(-1,2)\quad$ (b) $(-1,1)\quad$ (c) $(1,2)\quad$ (d) $(-2,-1)$

My Approach : I took various cases . First of all I took $a$ as positive and tried to solve it using various graphs of this equation . But still not getting appropriate answer . I also tried it by verifying the options . Please tell me the correct approach for doing this question .

Answer 1

If $a>0,4a+3b+2c>0\iff2c>-4a-3b$

$$0=2(ax^2+bx+c)>2ax^2+2bx-4a-3b$$

Now the roots of $$2ax^2+2bx-4a-3b=0$$ are $$\dfrac{-b\pm\sqrt{b^2+2a(4a+3b)}}{2a}$$

Now $b^2+2a(4a+3b)=(b+3a)^2-a^2<(b+3a)^2$

If $b+3a\ge0,\dfrac{-b+\sqrt{b^2+2a(4a+3b)}}{2a}\le\dfrac{-b+b+3a}{2a}=\dfrac32$

So, in that case a root lies near $\dfrac32$ which lies $\in\left[1,2\right]$

Answer 2

Since $b=-a(x_1+x_2)$ and $c=ax_1x_2$ we have to check when is $$a(4a+3b+2c) = a^2(4-3(x_1+x_2)+2x_1x_2)>0$$ so when $$4-3(x_1+x_2)+2x_1x_2>0$$ true, which is not difficult to check.

You can also rewrite last inequality like this: $$(2x_1-3)(2x_2-3)>1$$ Here is a graph: