<s>[INST] Quadratic equation with parameters If $a$, $b$, $c$ are real numbers such that $2a + 3b + 6c = 0$, prove that $ax^2+bx +c=0$ has a solution in the interval $[0, 1]$.
This should use high school maths or a little bit more then that.
 [/INST] \begin{align*}
  0 &= 2a+3b+6c \\
  f(x) &= ax^2+bx+c \\
  f(0) &= c \\
  f\left( \frac{1}{2} \right) &= \frac{a}{4}+\frac{b}{2}+c \\
  &= -\frac{a}{12} \\
  f(1) &= a+b+c \\
  &= \frac{a}{3}-c
\end{align*}
Case I: $ac \ge 0$
$$f(0) f\left( \frac{1}{2} \right) = -\frac{ac}{12} \le  0$$


*

*$\exists x\in \left[ 0, \dfrac{1}{2} \right]$ such that $f(x)=0$.


Case II: $ac \le 0$
$$f\left( \frac{1}{2} \right) f(1)= -\frac{a^2}{36}+\frac{ac}{12} \le  0$$


*

*$\exists x\in \left[ \dfrac{1}{2}, 1 \right]$ such that $f(x)=0$.



Combining, $\exists x \in [0,1]$ such that $f(x)=0$

N.B.
When $0 <3ac <a^2$, there're two such roots, namely $\alpha \in \left( 0, \frac{1}{2} \right)$ and  $\beta \in \left( \frac{1}{2}, 1 \right)$.
Updates
The value of $x=\dfrac{1}{2}$ can be inspired by plotting a family of curves by varying either $b$ or $c$.  You can see a fixed point at $x=\dfrac{1}{2}$ by varying $c$ below.

Alternatively, $(x,y)=\left(\dfrac{1}{2},-\dfrac{a}{12} \right)$ is a solution of
$$ax^2-\left( \frac{2a}{3} + 2c \right)x+c-y=
\frac{\partial}{\partial c}
\left[
  ax^2-\left( \frac{2a}{3} + 2c \right)x+c-y
\right]=0$$
</s>