First slide
Theory of equations
Question

If P(x)=ax2+bx+c and Q(x)=-ax2+dx+c where ac0, then P(x).Q(x)=0 has at least

Moderate
Solution

Let all four roots are imaginary. Then roots of both equations P(x)=0 and Q(x)=0 are imaginary.

Thus b2-4ac<0; d2+4ac<0, so b2+d2<0, which is impossible unless b=0, d=0.

So, if b0 or d0 at least two roots must be real.

If b=0, d=0, we have the equations.

P(x)=ax2+c=0 and Q(x)=-ax2+c=0

or x2=-ca; x2=ca as one of ca and -ca must be positive, so two roots must be real.

Get Instant Solutions
When in doubt download our app. Now available Google Play Store- Doubts App
Download Now
Doubts App