x2+a2nā1x>0āāxā 0 Which is not always true Method II: P(x)>x2 P(x)=x2+f(x)āāāāāf(x)>0āāxāR0 āµ P(0)=0āf(0)=0 Now, Pā(x) = 2 + fā(x) fā(0) = negative ā f(x) is concave down so f(x) canāt be +ve always ā which is contradiction