Let $a,b,c \in \mathbb R$ such that no two of them are equal and satisfy $$\det\begin{bmatrix}2a&b&c\\b&c&2a\\c&2a&b\end{bmatrix} = 0 ,$$ then the equation $24ax^2 + 4bx +c=0$ has:
a) atleast one root in $[0,\frac 12]$
b) at least one root in $[-\frac 12, 0)$
c) at least one root in $[1,2]$
d) none of these
On solving the determinant equation I have obtained $2a+b+c =0$ but I am not sure how that is of any help. because I am uanble to transform the given equation in $x$ to $2a+b+c= 0$ for any $x$. How do I proceed?
Edit:
Using @AnotherJohnDoe's suggestion, if we let $c=0$, we get $b=-2a$ after solving the determinant equation.
On substituting these values into the quadratic equation, we get:
$-12bx^2 + 4bx = 0$
which has roots $\dfrac 13$ and $0$ so we get option A/D. But there doesn't seem to be a way to eliminate option D. So there must be another solution to this question?
Let $f(x)=24ax^2+4bx+c$
Suppose $f(0)\gt0$. Then it follows $c\gt0$. Now, $f(\frac 12)=6a+2b+c=2a-c$. If $f(\frac 12)\gt0,$ then $2a\gt c\gt 0.$
Then $f(\frac 14)=\frac{24}{16}a+b+c=-\frac 12a<0.\\$
Similarly, if $f(0)\lt 0$, then $c\lt 0$. If $f(\frac 12)=2a-c\lt0$, then $2a\lt c\lt 0\Rightarrow f(\frac 14)=-\frac 12\gt 0$.