<s>[INST] Prove that $ ax^2+bx+c=0 $ has at least one root in $(0,1)$ if $10a+12b+15c=0$ 
If $10a+12b+15c=0$, Prove that $$ ax^2+bx+c=0 $$ has at least one root in $(0,1)$.

Progress
I tried to solve this by Rolle`s theorem ($f'$ has a root between any two roots of $f$), but could not reach the result.
 [/INST] Consider $x = \frac{10}{12}$:
$$\begin{align*}
f\left(\frac{10}{12}\right)
&= a\left(\frac{10}{12}\right)^2+b\left(\frac{10}{12}\right)+c\\
&= \frac{10}{144}\left(10a+12b+\frac{144}{10}c\right)\\
&= \frac{10}{144}\left(10a+12b+15c-\frac{6}{10}c\right)\\
&= -\frac{6}{144}c
\end{align*}$$
And consider $x=0$:
$$f(0) = 0^2a + 0b + c = c$$
If $c\ne 0$, by intermediate value theorem, there must be an $x\in\left(0, \frac{10}{12}\right)\subset(0,1)$ which satisfies $f(x)= 0$.
If $c=0$, $\frac{10}{12}\in(0,1)$ is a root.

How I obtained $x = \frac{10}{12}$:
I assume $x = 12k$ and $x^2 = 10k$ for some real number $k$. By solving $10k = (12k)^2$, $k$ is chosen to be the non-zero root $k = \frac{10}{144}$, which appears as the multiplier above. And $x = 12k = \frac{10}{12}$.
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