If $ax^2+bx+c=0,a\neq 0$ has two distinct roots,then $a(5a+2b+c)$ is

Question

If $ax^2+bx+c=0,a\neq 0,a,b,c\in \mathbb{R}$ has two distinct real roots in $(1,2)$,then $a(5a+2b+c)$ is

$(A)$positive $\hspace{1cm}(B)$negative $\hspace{1 cm} (C)$zero $\hspace{1 cm}(D)$none of these

My attempt:I tried $b^2-4ac>0$. I tried $f(1)>0,f(2)>0$ and $f(1)<0,f(2)<0$ but could not reach desired expression. Can someone guide me?

Answer 1

As you noted, if $a > 0$ then $f(1) > 0$ and $f(2) > 0$. Also, if $a < 0$ then $f(1) < 0$ and $f(2) < 0$.

Now, write $a(5a+2b+c) = a(a+4a+2b+c) = a(a+f(2))$.

For both cases $a > 0$ and $a < 0$, what can you say about $a(a+f(2))$?

Answer 2

Hint. Let $f(x)=ax^2+bx+c$. Then $a(4a+2b+c)=af(2)>0$.

Answer 3

Do you know that :

$- b/a = r_1 + r_2 = : sum$ where $r_1$ and $r_2$ are distinct roots.

$ c/a = r_1 \times r_2 = : prod$

Let's estimate upper and lower limits of $sum$ and $prod$.

$ 1+1 < sum < 2+2$ and $1\times 1 < prod < 2 \times 2$.

Thus : $ a (5a + 2b + c) = a^2(5 - 2 \times sum + prod)$

is max when sum is min and prod is max and viceversa for min.

Evaluating : upper bound is $a^2(5-2 \times 1 + 4) = 7a^2$ and

lower bound is $a^2(5-2\times4+4) = a^2$. As $a \neq 0$,

the expression is definitely positive and between above two values.