First slide
Theory of equations
Question

If a(1,1) then roots of the quadratic equation (a1)x2+ax+1a2=0 are 

Moderate
Solution

(a1)x2+ax+1a2=0 D=a24(a1)1a2 =a24a1a2+41a2    a21a22+41a211a2>0 for a(1,1) 

Hence roots are real but not equal.

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