Prove that for any $f_1,f_2,...f_k \in I$, there exists a point $x_0 \in [a,b]$ such that $f_1(x_0)=...=f_k(x_0)=0$. Let $C[a,b]$ be the ring of real-valued functions continuous on $[a,b]$ and let $I \subset C[a,b]$ be its proper ideal. Prove that for any $f_1,f_2,...f_k \in I$, there exists a point $x_0 \in [a,b]$ such that $f_1(x_0)=...=f_k(x_0)=0$.
My attempt: Suppose not. Then we have $f_i(x) \neq 0$ for all $x \in [a,b]$. Suppose $f>0$. Define $g(x)=\sum_{i=1}^k{f_i(x)}$. Clearly $g>0$ for all $x \in [a,b]$. 
I have a feeling that I need to apply the condition of proper ideal somewhere but I am unable to see it. Can anyone guide me?
 A: Hint
I will show it for two functions and then the idea can be generalized. Suppose $\not\exists \, x_0 \in [a,b]$ such that $f_1(x_0)=0=f_2(x_0)$, i.e they do not both vanish at the same point then consider the function $h(x)=(f_1(x))^2+(f_2(x))^2$ in the ideal $I$. Now corresponding to this $h$, we will have the function 
$$g(x)=\frac{1}{(f_1(x))^2+(f_2(x))^2}.$$
This will definitely be a continuous function on $[a,b]$ and the denominator doesn't vanish (by our hypothesis). Now by absorption property of ideal $gh =1 \in I$. But once $1 \in I$, this will show that $I=C[a,b]$, so $I$ is not proper ideal.
A: Hint : if this is not the case, take $f_1, ..., f_k$ such that for every $x \in [a,b]$, there is a $i \leq k$ such that $f_i (x) \neq 0$. Define a function $g$ by $g = f_1^2 + ... + f_k^2$. What could you say about $g$ ? 
Answer. As a consequence of the fact that there is no $x \in [a,b]$ such that all the chosen $f_i$'s are zeros, the function $g$ never takes the value $0$, so it a unit in $\mathcal{C}[a,b]$. But it also belongs to the ideal $I$, by construction. So $I$ is a proper ideal containing a unit, which is absurd. 
