If one root of the cubic equation $x^3+36=7x^2$ is double of another, then the number of negative roots are

Let a, 2a and b be roots of the will cubic equation $x^3+36-7x^2=0$

Sum of roots $=-\frac{\left(-7\right)}{1}$

$a+2a+b=7$

$3a+b=7\to \left(1\right)$

$S_3=\frac{-d}{a}$

$a\left(2a\right)b=-36$

$2a^{2}b=-36$

$S_2=0$

$ab+a\left(2a\right)+2a\left(b\right)=0$

$2a^2+3ab=0\Rightarrow a\left(2a+3b\right)=0$

$b=\frac{-2a}{3}$

$3a-\frac{2a}{3}=7\Rightarrow 7a=21$$a=3$$b=-2$

$\therefore$ roots are 3,6,-2