First slide
Theory of equations
Question

 If f(x)=ax2+bx+c,g(x)=ax2+bx+c where ac0 then f(x)g(x)=0 has 

Moderate
Solution

 I et D1 and D2 be discriminants of ax2+bx+c0 and -ax2+bx+c=0 respectively 

 Then, D1=b24ac,D2=b2+4ac

 Now,  ac0 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. 

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