if x2-x+a-3 0 for atleast one negative value of x then complete set of values for a are a (-infinity,4) - Maths - Logs Equations and Inequalities

Question

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)

Rahul Raj · Accepted Answer

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

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)

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)