So the questions says -
Let $f(x), g(x)$ and $h(x)$ be quadratic polynomials having positive leading coefficients and real and distinct roots. If each pair of them has a common root, then find roots of $f(x)+g(x)+h(x)=0$.
What I did -
Let,
$$ f(x) = a_1 (x-\alpha) (x-\beta), \\ g(x) = a_2 (x-\beta)(x-\gamma), \\ h(x) = a_3 (x-\gamma) (x-\alpha), \\ F(x):=f(x)+g(x)+h(x)$$ Now,
$$ F(\alpha) = a_2 (\alpha-\beta) (\alpha-\gamma) \\ F(\beta) = a_3 (\beta-\gamma) (\beta-\alpha) \\ F(\gamma) = a_1 (\gamma-\alpha) (\gamma-\beta)$$
I don't know how to proceed further. I referred to the solution, it just multiplies $F(\alpha), F(\beta) \text{, and } F(\gamma)$ and it comes out to be negative. And hence it concludes that roots of $F(x)=0$ are real and distinct. Can anyone explain why?
Thanks.
Suppose, without loss of generality, that $\alpha<\beta<\gamma$: it is easy to check that $F(\alpha)>0$, $F(\beta)<0$ and $F(\gamma)>0$. But $F(x)$ is a quadratic polynomial, hence a continuous function: it follows that $F(x)=0$ for some $x$ between $\alpha$ and $\beta$, and also for some $x$ between $\beta$ and $\gamma$.