if x2-x+a-3<0 for atleast one negative value of x then complete set of values for a are a.(-infinity,4) b.(-infinity,2) c.(-infinity,3) d.(-infinity,1) Share with your friends Share 0 Rahul Raj answered this Dear student x2-x+a-3=0if roots are α and βα+β=--11=1αβ=a-31=a-3the coefficient of x2 is positive, so the equation is concave up and will be negative only between the rootsas one root has to be negativeproduct of roots is negative (as sum is positive)a-3<0a<3a∈(-∞,3) Regards 0 View Full Answer