If $3(a+2c) =4(b+3d)$ then the equation $a x^3 + b x^2 + c x +d =0$ will have real roots in which interval?

Question

I tried some vales to satisfy the equality and conjectured that atleast one root will lie in the interval $[-1,0]$. But couldn't prove. Please help

Answer 1

Let $P(X)=3aX^4+4bX^3+6cX^2+12dX$.

$P(-1)=P(0)=0$ hence $P'(x)=0$ for some $x\in[-1,0]$.

Answer 2

Idea: Replace $a\to 4a$, $b\to 3b$ and $c\to 2c$ then we get $d= a-b+c$ and consider equation

$$0 = 4ax^3+3bx^2+2cx+a-b+c$$ because of homogenity we can assume that $a=1$ so we have to consider: $$p(x) = 4x^3+3bx^2+2cx+1-b+c$$ Now it is enough to prove that two among $p(0)$, $p(-{1\over 2})$ and $p(-1)$ have different sign.