Let $a, 2 a$ and $b$ be roots of the cubic equation $x^3+36-7 x^2=0$
So, sum of roots $=-\frac{(-7)}{1}$
$\therefore \quad a+2 a+b=7$
$\Rightarrow \quad 3 a+b=7$ ...(i)
Now, $a(2 a) b=36$ $\left[\right.$ product of roots $\left.=\frac{d}{a}\right]$
$2 a^2 b=36$ and $a b+a(2 a)+2 a(b)=0$
$\begin{aligned} \Rightarrow & & 2 a^2+3 a b & =0 \Rightarrow a(2 a+3 b)=0 \\ \Rightarrow & & b & =\frac{-2 a}{3}\end{aligned}$
$\Rightarrow \quad 3 a-\frac{2 a}{3}=7 \quad$ [from Eq. (i)]
$\Rightarrow \quad \frac{9 a-2 a}{3}=7 \Rightarrow 9 a-2 a=21$
$\Rightarrow \quad 7 a=21 \Rightarrow a=3$
and $b=7-3 a$
$=7-3(3)=-2$ [from Eq. (i)]
Roots are $3,6,-2$
$\therefore$ Number of negative root is 1 .