Prove that if $12a+6b+4c+3d=0$, then $a+bx+cx^2+dx^3=0$ has a real solution in $(0,1)$

It turns out that $12a+6b+4c+3d=0$ is the condition that makes Rolle's Theorem applicable. Let $$ f(x)=\int_0^x(a+bt+ct^2+dt^3)dt=ax+\frac b2x^2+\frac b3x^3+\frac d4x^4.$$ Then clearly $f(x)$ is continuous in $[0,1]$, differentiable in $(0,1)$, $f(0)=0$, and $$f(1)=a+\frac b2+\frac b3+\frac d4=\frac{1}{12}(12a+6b+4c+3d)=0.$$ By Rolle's Theorem, there is $\xi\in(0,1)$ such that $f'(\xi)=0$, namely $a+bx+cx^2+dx^3=0$ has a root in $(0,1)$.

We know Every cubic equation $dx^3+cx^2+bx+a=0$ with real coefficients and $d\not= 0$ has three solutions (some of which may equal each other if they are real, and two of which may be complex non-real numbers) and at least one real solution. So now we put $d=0$ and we get : $$cx^2+bx+a=0$$ and $$12a + 6b+ 4c =0 \to 6a+3b+2c = 0 \to b = \frac{6a+2c}{-3}$$ Now we want to $\Delta_1 \ge0$ ( to has real root).

$\Delta_1 = b^2-4ac = \frac{36a^2+4c^2+24ac}{9} - 4ac = \frac{36a^2+4c^2-12ac}{9}\ge0\iff 9a^2-3ac+c^2\ge0$

Which it is obvious because $\Delta_2 = -27a^2c^2\le0$

Note : When $\Delta \lt 0 $ quadratic polynomial has same sign as coefficient of $x^2$.(In this case 9) and if $\Delta = 0$ value of quadratic polynomial will be zero at one point and in other points has same sign as coefficient of $x^2$.