If `f(x)=a x^2+b x+c ,g(x)=-a x^2+b x+c ,w h e r ea c!=0,` then prove that `f(x)g(x)=0` has at least two real roots.

Explanation

Let `D_(1) and D_(2)` be discriminants of `ax^(2) + bx + c =0 and - ax^(2) +bx + c = 0`, respectively. Then
`D_(1) = b^(2) - 4ac,D_(2) = b^(2) + 4ac.`
Now, ac`ne 0 rArr` either ac `gt` 0 or ac `lt` 0
If ac`gt`, then`D_(2) gt 0`. Therefore, roots of `-ax^(2) + bx + c = 0` are real.
If ac`lt`0, then `D_(1) gt 0`. Therefore, roots of `ax^(2) + bx + c = 0` are real. Thus , `f(x)g(x) = 0` has at least two real roots.