If f(x)=ax2+bx+c,g(x)=−ax2+bx+c where ac≠0 then f(x)g(x)=0 has
at least two real roots
two real roots and two imaginary roots
at least three real roots
no real roots
Let D1 and D2 be discriminants of ax2+bx+c=0 and −ax2+bx+c=0 respectively.
Then, D1=b2−4ac,D2=b2+4ac
Now, ac≠0⇒ either ac>0 or ac<0
If ac>0, then D2>0. Therefore, roots of −ax2+bx+c=0 are real
If ac<0, then D1>0. Therefore, roots of ax2+bx+c=0 are real
Thus, f(x)g(x) has at least two real roots.