Quadratic Equation Problem - ?

Question

Prove that If 0 is less than or equal to p which itself is less than or equal to π , then the quadratic eqn

( cos p - 1 ) x^2 + ( cos p ) x + sin p = 0 has real roots.

97/50

Answer 1

The discriminant ‘ D ‘  of QE ( quadratic Equation )  

D  =  cos^2 p – 4 ( cos p – 1 ) sin p

=>   cos^2 p  Ã¢Â€Â“  4 cos p . sin p  +  4 sin p

=>  ( cos p – 2 sin p )^2  +  4 sin p  -  4  sin^2  p

=>    ( cos p – 2 sin p )^2  +  4 sin p  (  1 – sin p )

As  0 is less than or equal to p which itself is less than or equal to π  

and also  (  1  -  sin p  )  is greater than or equal to , p ε R. Therefore,  

D is positive. Hence the given equation has real roots. ……………….. Proved

Dr. P.K.T

Answer 2

The problem is this is too much math for me!