First slide
Theory of equations
Question

If ax2+(bc)x+abc=0 has unequal real roots for all cR, then

Moderate
Solution

We have,

 D=(bc)24a(abc)>0 or  b2+c22bc4a2+4ab+4ac>0 or  c2+(4a2b)c4a2+4ab+b2>0 for all cR

Discriminant of the above expression in c must be negative.

Hence,

    (4a2b)244a2+4ab+b2<0 or     4a24ab+b2+4a24abb2<0 or     a(ab)<0     a<0 and ab>0 or a>0 and ab<0     b<a<0 or b>a>0

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